Light OJ Beginners Problems

#include <cstdio>
#include <iostream>
using namespace std;
int main()
{int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int a=0,b=n;if(n>10)a=10,b=n-10;printf("%d %d\n",a,b);}return 0;
}

构造矩阵

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
const long long mod=10000007;
using namespace std;
struct Matrix
{long long m[7][7];//之前用的int,WA了，然后__int64 PE 了
};
Matrix E,B;
int a[7];
void Init()
{for(int i=1;i<=6;i++)for(int j=1;j<=6;j++){E.m[i][j]=(i==j);B.m[i][j]=0;}for(int i=1;i<=6;i++)B.m[i][1]=1;for(int i=1;i<=5;i++)B.m[i][i+1]=1;
}
Matrix Multi(Matrix A,Matrix B,int M,int N,int K)
{Matrix ans;for(int i=1;i<=M;i++)for(int j=1;j<=N;j++){ans.m[i][j]=0;for(int k=1;k<=K;k++){ans.m[i][j]=(ans.m[i][j]+((A.m[i][k]%mod)*(B.m[k][j])%mod))%mod;}  // 我去，WA了三次，就感觉这里有问题，尼玛，改成longlong就过了，__int64会PE}return ans;
}
Matrix Pow(Matrix A,int k)
{Matrix ans=E;while(k){if(k&1){k--;ans=Multi(ans,A,6,6,6);}else{k/=2;A=Multi(A,A,6,6,6);}}return ans;
}
int main()
{int t;scanf("%d",&t);Init();for(int ii=1;ii<=t;ii++){int n;Matrix D;for(int i=1;i<=6;i++){scanf("%d",&a[i]);D.m[1][7-i]=a[i];}scanf("%d",&n);printf("Case %d: ",ii);if(n<6)printf("%d\n",a[n+1]%mod);else{n-=5;Matrix ans=Multi(D,Pow(B,n),1,6,6);printf("%lld\n",ans.m[1][1]%mod);}}return 0;
}

找规律

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
long long t,n;
long long x,y;
void solve()
{long long cnt=sqrt(n*1.0);if(cnt*cnt<n)cnt++;n-=(cnt-1)*(cnt-1);if(cnt&1){if(n>=cnt){y=cnt;x=n-cnt;x=cnt-x;}else{x=cnt;y=n;}}else{if(n>=cnt){x=cnt;y=n-cnt;y=cnt-y;}else{y=cnt;x=n;}}
}
int main()
{scanf("%lld",&t);for(long long i=1;i<=t;i++){scanf("%lld",&n);solve();printf("Case %lld: %lld %lld\n",i,x,y);}return 0;
}

//我去  WA了那么多次。。。。
#include <cstdio>
#include <iostream>
using namespace std;
int solve(int m,int n)
{if(m==1||n==1)return m*n;if(m==2||n==2){int ans=m*n/8*4;if((m*n)%8==2)return ans+=2;if((m*n)%8==6||(m*n)%8==4)return ans+=4;}return (m*n+1)/2;
}
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int a,b;scanf("%d%d",&a,&b);printf("Case %d: ",i);printf("%d\n",solve(a,b));}return 0;
}

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int n;scanf("%d",&n);int ans=0,a;for(int j=1;j<=n;j++){scanf("%d",&a);if(a>0)ans+=a;}printf("Case %d: %d\n",i,ans);}return 0;
}

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
using namespace std;
#define PI (2*acos(0.0))
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){double n;scanf("%lf",&n);printf("Case %d: %.2lf\n",i,4*n*n-n*n*PI);}return 0;
}

好题 位运算 STL

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int fun(string a)
{int ans=1;for(int i=1;i<a.size();i++){if(a[i]=='0')ans*=2;elseans=(ans*2+1);}return ans;
}
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int n;scanf("%d",&n);string str="";while(n){if(n&1)str.insert(0,"1");elsestr.insert(0,"0");n/=2;}if(!next_permutation(str.begin(),str.end())){str.insert(0,"10");str.erase(str.size()-1,1);}printf("Case %d: %d\n",i,fun(str));}return 0;
}

math

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
#define maxn 1000001
double a[maxn];
int main()
{a[0]=0;for(int i=1;i<=maxn;i++)a[i]=a[i-1]+log(i+0.0);int t;scanf("%d",&t);for(int i=1;i<=t;i++){int m,n;scanf("%d%d",&m,&n);printf("Case %d: %.lf\n",i,floor(a[m]/log(n*1.0)+1));}return 0;
}

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
long long a[4];
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){for(int j=1;j<=3;j++)scanf("%lld",&a[j]);sort(a+1,a+4);printf("Case %d: ",i);if(a[1]*a[1]+a[2]*a[2]==a[3]*a[3])printf("yes\n");elseprintf("no\n");}return 0;
}

#include <cstdio>
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int a,b;scanf("%d%d",&a,&b);int ans=a-b;ans>0?ans:ans*=-1;ans+=a;ans*=4;ans+=19;printf("Case %d: ",i);printf("%d\n",ans);}return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define PI 3.14159265
using namespace std;
int t;
int main()
{scanf("%d",&t);for(int i=1;i<=t;i++){double R;int n;scanf("%lf%d",&R,&n);double r=R*sin(PI/n)/(1+sin(PI/n));printf("Case %d: %lf\n",i,r);}return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int x1,y1,x2,y2;scanf("%d%d%d%d",&x1,&y1,&x2,&y2);int n;scanf("%d",&n);printf("Case %d:\n",i);while(n--){int a,b;scanf("%d%d",&a,&b);if(a>x1&&a<x2&&b>y1&&b<y2)printf("Yes\n");elseprintf("No\n");}}return 0;
}

算约数个数 排序

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 1001
struct S
{int a,b;
};
S s[maxn];
int prime[maxn];
void Prime()
{memset(prime,1,sizeof(prime));prime[0]=prime[1]=0;for(int i=2;i*i<=maxn;i++)if(prime[i])for(int j=2*i;j<=maxn;j+=i)prime[j]=0;
}
int tmp[maxn];
bool cmp(S x,S y)
{if(x.a<y.a)return true;if(x.a==y.a)return x.b>y.b;return false;
}
int main()
{Prime();for(int i=1;i<=maxn;i++)tmp[i]=1;for(int i=1;i<=maxn;i++){for(int j=2;j<=i;j++)if(prime[j]&&i%j==0){int cnt=1;int ii=i;while(ii%j==0){ii/=j;cnt++;}tmp[i]*=cnt;}}for(int i=1;i<maxn;i++){s[i].a=tmp[i];s[i].b=i;}sort(s+1,s+maxn,cmp);int t;scanf("%d",&t);for(int i=1;i<=t;i++){int n;scanf("%d",&n);printf("Case %d: %d\n",i,s[n].b);}return 0;
}

栈

#include <cstdio>
#include <iostream>
#include <string>
#include <stack>
#include <algorithm>
using namespace std;
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){stack<string>F,B;string str="http://www.lightoj.com/",command;printf("Case %d:\n",i);while(cin>>command&&command!="QUIT"){if(command=="BACK"){if(B.empty())printf("Ignored\n");else{F.push(str);str=B.top();B.pop();cout<<str<<endl;}}else if(command=="FORWARD"){if(F.empty())printf("Ignored\n");else{B.push(str);str=F.top();F.pop();cout<<str<<endl;}}else{B.push(str);cin>>str;cout<<str<<endl;while(F.size())F.pop();}}}return 0;
}

#include <cstdio>
#include <iostream>
using namespace std;
int t;
int main()
{scanf("%d",&t);for(int i=1;i<=t;i++){long long n;scanf("%lld",&n);printf("Case %d: ",i);if(n&1){printf("Impossible\n");}else{long long ans=1;while(n%2==0){ans*=2;n/=2;}printf("%lld %lld\n",n,ans);}}return 0;
}

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int a[110];
char str[10];
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){int n,m;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)scanf("%d",&a[i]);while(m--){scanf("%s",str);if(str[0]=='P'){int x,y;scanf("%d%d",&x,&y);x++;y++;swap(a[x],a[y]);}else if(str[0]=='R'){for(int i=1;i<=n/2;i++)swap(a[i],a[n+1-i]);}else if(str[0]=='S'){int num;scanf("%d",&num);for(int i=1;i<=n;i++)a[i]+=num;}else if(str[0]=='M'){int num;scanf("%d",&num);for(int i=1;i<=n;i++)a[i]*=num;}else{int num;scanf("%d",&num);for(int i=1;i<=n;i++)a[i]/=num;}}printf("Case %d:\n",ii);for(int i=1;i<n;i++)printf("%d ",a[i]);printf("%d\n",a[n]);}return 0;
}

分治

/*
// 分治
// 先求从1到n的个数
//枚举之后找规律
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[60];
int b[60];
int main()
{int ans=0;memset(b,0,sizeof(b));for(int i=1;i<=9;i++){ans+=i%10;ans%=3;if(ans%3==0)b[i]=b[i-1]+1;elseb[i]=b[i-1];}for(int i=10;i<=50;i++){int ii=i;ans+=ii%10;ii/=10;ans+=ii;ans%=3;if(ans%3==0)b[i]=b[i-1]+1;elseb[i]=b[i-1];}for(int i=1;i<=50;i++)printf("%d.......%d\n",i,b[i]);return 0;
}
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int fun(int n)
{if(n==1||n==0)return 0;return n-1-(n-1)/3;
}
int main()
{int a,b,t;scanf("%d",&t);for(int i=1;i<=t;i++){scanf("%d%d",&a,&b);printf("Case %d: %d\n",i,fun(b)-fun(a-1));}return 0;
}

// 之前看那个位运算的时候发现了这篇博客
// http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=bitManipulation
// 用了下 __builtin_popcount() 感觉还可以
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int n;scanf("%d",&n);printf("Case %d: ",i);if(__builtin_popcount(n)&1)printf("odd\n");elseprintf("even\n");}return 0;
}

// 先打表看了一下，18的阶乘就是一个18位数
// 所以就算是枚举也不算难
// 但是这个题可以贪心直接做，因为前n-1项的和肯定是小于第n项的
#include <cstdio>
#include <iostream>
#include <stack>
using namespace std;
long long fac(long long n)
{if(n==1||n==0)return 1;return n*fac(n-1);
}
long long f[20];
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){long long n;scanf("%lld",&n);printf("Case %d: ",i);stack<int>s;while(!s.empty())s.pop();int cnt=20;int tag=0;while(n){if(cnt<0){tag=1;break;}if(n>=fac(cnt)){n-=fac(cnt);s.push(cnt);}cnt--;}if(tag)printf("impossible\n");else{while(s.size()!=1){printf("%d!+",s.top());s.pop();}printf("%d!\n",s.top());}}return 0;
}

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int x1,y1,x2,y2;scanf("%d%d%d%d",&x1,&y1,&x2,&y2);printf("Case %d: ",i);x1=abs(x1-x2);y1=abs(y1-y2);if(x1==y1)printf("1\n");else{if(x1%2==y1%2)printf("2\n");elseprintf("impossible\n");}}return 0;
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int n;int x1,x2,y1,y2,z1,z2;int x=0,y=0,z=0;int X=1001,Y=1001,Z=1001;scanf("%d",&n);printf("Case %d: ",i);for(int j=1;j<=n;j++){scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);x=max(x,x1);y=max(y,y1);z=max(z,z1);X=min(x2,X);Y=min(y2,Y);Z=min(z2,Z);}if(X>x&&Y>y&&Z>z)printf("%d\n",(X-x)*(Y-y)*(Z-z));elseprintf("0\n");}return 0;
}

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <deque>
using namespace std;
string str;
int num;
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){printf("Case %d:\n",i);int m,n;scanf("%d%d",&n,&m);deque<int>q;while(m--){cin>>str;if(str=="pushLeft"){cin>>num;if(q.size()!=n){q.push_back(num);cout<<"Pushed in left: "<<num<<endl;}elsecout<<"The queue is full"<<endl;}else if(str=="pushRight"){cin>>num;if(q.size()!=n){q.push_front(num);cout<<"Pushed in right: "<<num<<endl;}elsecout<<"The queue is full"<<endl;}else if(str=="popRight"){if(q.empty())cout<<"The queue is empty"<<endl;else{cout<<"Popped from right: "<<q.front()<<endl;q.pop_front();}}else{if(q.empty())cout<<"The queue is empty"<<endl;else{cout<<"Popped from left: "<<q.back()<<endl;q.pop_back();}}}}return 0;
}

大数整除否

// 我去。。套了模板样例最后一个还是过不了
//  算了 还是看大神的做法
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;
bool is_ok(vector<long long>a,int b)
{for(int i=a.size()-1;i>=0;i--){long long cnt=(a[i]%b)*10;if(cnt!=0&&i==0)return false;if(i!=0)a[i-1]+=cnt;}return true;
}
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){vector<long long>a;string aa;int b;cin>>aa;cin>>b;printf("Case %d: ",ii);for(int i=aa.size()-1;i>=0;i--)if(aa[i]!='-')a.push_back(aa[i]-'0');if(is_ok(a,b))printf("divisible\n");elseprintf("not divisible\n");}return 0;
}

算圆锥体积

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#define PI (2*acos(0.0))
using namespace std;
int main()
{int cases;cin>>cases;for(int ii=1;ii<=cases;ii++){double r1,r2,p,h;cin>>r1>>r2>>h>>p;cout<<"Case "<<ii<<": ";double ans,x,v;x=h*r2/(r1-r2);ans=(r1-r2)/h*(p+x);v=1.0/3*PI*ans*ans*(p+x)-1.0/3*PI*r2*r2*x;printf("%.8lf\n",v);}return 0;
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
char s[20];
char str[20];
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){scanf("%s",s);printf("Case %d: ",i);strcpy(str,s);//strrev(str);  我去，用这个PE了for(int i=0;i<strlen(str)/2;i++)swap(str[i],str[strlen(str)-1-i]);if(strcmp(s,str)==0)printf("Yes\n");elseprintf("No\n");}return 0;
}

贪心

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[40];
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){int n,p,q;scanf("%d%d%d",&n,&p,&q);for(int i=1;i<=n;i++)scanf("%d",&a[i]);sort(a+1,a+n+1);printf("Case %d: ",ii);int ans=0;int cnt=1;q-=a[cnt];while(q>=0){ans++;if(cnt==n) //这个地方之前忽略了，WA了break; // 细节问题要重视啊cnt++;q-=a[cnt];if(ans>=p)break;}printf("%d\n",ans);}return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[20];
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);int ans=0;a[0]=2;//我去，漏掉这个了，看了好久的forum，之前当0做的for(int i=0;i<=n-1;i++){if(a[i+1]>a[i])ans+=(a[i+1]-a[i]-1)/5+1;}printf("Case %d: %d\n",ii,ans);}return 0;
}

结构体排序

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
struct stu
{char name[25];int sum;
};
bool cmp(stu x,stu y)
{return x.sum<y.sum;
}
stu s[101];
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){int n;scanf("%d",&n);for(int i=1;i<=n;i++){int x,y,z;scanf("%s%d%d%d",s[i].name,&x,&y,&z);s[i].sum=x*y*z;}printf("Case %d: ",ii);sort(s+1,s+n+1,cmp);if(s[1].sum==s[n].sum)printf("no thief\n");else{printf("%s took chocolate from %s\n",s[n].name,s[1].name);}}return 0;
}

这个做到想吐

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{int t,num;cin>>t;for(int ii=1;ii<=t;ii++){cout<<"Case "<<ii<<": ";int n,m,k,p;cin>>n>>m>>k;vector<int>v1(m+1);vector<vector<int> >v2(n+1);for(int i=1;i<=n;i++){v2[i].push_back(0);for(int j=1;j<=k;j++){cin>>num;v2[i].push_back(num);}}cin>>p;for(int i=1;i<=p;i++){cin>>num;v1[num]=1;}bool tag=0;for(int i=1;i<=n;i++){int cnt=0;for(int j=1;j<=k;j++){if(v2[i][j]<0&&!v1[-v2[i][j]])cnt++;if(v2[i][j]>0&&v1[v2[i][j]])cnt++;}if(cnt==0)tag=1;}if(!tag)cout<<"Yes\n";elsecout<<"No\n";}return 0;
}

#include <cstdio>
#include <iostream>
using namespace std;
int main()
{long long a,b;int t;scanf("%d",&t);for(int i=1;i<=t;i++){scanf("%lld%lld",&a,&b);printf("Case %d: ",i);printf("%lld\n",a/2*b);}return 0;
}

平行四边形面积

#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
struct LPoint
{double x,y;
};
struct Ldir//表示方向向量
{double dx,dy;
};
struct Line
{LPoint p;Ldir dir;
};
void format(Line l,double &A,double &B,double &C)
{A=l.dir.dy;B=-l.dir.dx;C=l.p.y*l.dir.dx-l.p.x*l.dir.dy;
}
double p2ldis(LPoint a,Line l)
{double A,B,C;format(l,A,B,C);return (fabs(A*a.x+B*a.y+C)/sqrt(A*A+B*B));
}
double Distance(LPoint a,LPoint b)
{return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){LPoint po[4];for(int i=1;i<=3;i++)scanf("%lf%lf",&po[i].x,&po[i].y);po[0].x=po[1].x+po[3].x-po[2].x;po[0].y=po[1].y+po[3].y-po[2].y;double ans=Distance(po[1],po[2]);Line l;l.p=po[1];l.dir.dx=po[2].x-po[1].x;l.dir.dy=po[2].y-po[1].y;ans*=p2ldis(po[0],l);printf("Case %d: ",ii);printf("%.lf %.lf %.lf\n",po[0].x,po[0].y,ans);}return 0;
}

#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{int tt;scanf("%d",&tt);for(int ii=1;ii<=tt;ii++){double v1,v2,v3,a1,a2;scanf("%lf%lf%lf%lf%lf",&v1,&v2,&v3,&a1,&a2);double t,s;s=v1*v1/2/a1+v2*v2/2/a2;t=max(v1/a1,v2/a2);printf("Case %d: %lf %lf\n",ii,s,t*v3);}return 0;
}

求面积

#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;
#define PI (2*acos(0.0))
double sita(double a,double b,double c)
{return acos((a*a+b*b-c*c)/(2*a*b));
}
double area(double a,double b,double c)
{double ans=(a+b+c)/2;return sqrt(ans*(ans-a)*(ans-b)*(ans-c));
}// 之前求三角形面积用角度求的，WA了，还是海伦公式丢失精度低
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){double r1,r2,r3;scanf("%lf%lf%lf",&r1,&r2,&r3);double a=r1+r2;double b=r2+r3;double c=r3+r1;double ans=area(a,b,c);ans-=sita(a,c,b)/2*r1*r1;ans-=sita(a,b,c)/2*r2*r2;ans-=sita(c,b,a)/2*r3*r3;printf("Case %d: %lf\n",ii,ans);}return 0;
}

统计字符串，看是否为子串

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstdlib>
using namespace std;
int main()
{int t;cin>>t;getchar();for(int ii=1;ii<=t;ii++){int a[26];int tag=1;memset(a,0,sizeof(a));char c;while(scanf("%c",&c),c!='\n'){if(c==' ') continue;if(c>='A'&&c<='Z') a[c-'A']++;if(c>='a'&&c<='z') a[c-'a']++;}while(scanf("%c",&c),c!='\n'){if(c==' '||tag==0) continue;if(c>='A'&&c<='Z'){if(a[c-'A']!=0)a[c-'A']--;elsetag=0;}else if(c>='a'&&c<='z'){if(a[c-'a']!=0)a[c-'a']--;elsetag=0;}}cout<<"Case "<<ii<<": ";if(tag)cout<<"Yes\n";elsecout<<"No\n";}return 0;
}

进制转换

#include <cstdio>
#include <stack>
#include <iostream>
#include <algorithm>
using namespace std;
int a[5];
char b[5][8];
int fun(stack<int>s)
{int ans=0;int dd=1;while(s.size()){ans+=dd*s.top();s.pop();dd*=2;}return ans;
}
int main()
{int t;int tmp;scanf("%d",&t);for(int ii=1;ii<=t;ii++){stack<int>s[5];scanf("%d.%d.%d.%d",&a[1],&a[2],&a[3],&a[4]);for(int i=1;i<=4;i++){for(int j=1;j<=8;j++){scanf("%1d",&tmp);s[i].push(tmp);}if(i!=4)scanf("%*c");}int tag=1;for(int i=1;i<=4;i++){if(fun(s[i])!=a[i]){tag=0;break;}}printf("Case %d: ",ii);if(tag)printf("Yes\n");elseprintf("No\n");}return 0;
}

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
char s[10];
int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){int ans=0,num;int n;printf("Case %d:\n",i);scanf("%d",&n);while(n--){scanf("%s",s);if(s[0]=='d'){scanf("%d",&num);ans+=num;}elseprintf("%d\n",ans);}}return 0;
}

// 直接做会TLE，已证明
//400年一个轮回，处理后再做
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
bool is_leap_year(int y)
{if(y%4==0){if(y%100==0){if(y%400==0)return true;return false;}return true;}return false;
}
int the_month(string a)
{if(a=="January")return 1;if(a=="February")return 2;if(a=="March")return 3;if(a=="April")return 4;if(a=="May")return 5;if(a=="June")return 6;if(a=="July")return 7;if(a=="August")return 8;if(a=="September")return 9;if(a=="October")return 10;if(a=="November")return 11;if(a=="December")return 12;elsereturn 0;
}
int main()
{int ii,t;cin>>t;for(ii=1;ii<=t;ii++){string str1,str2;int date1,date2;int year1,year2;cin>>str1;cin>>date1;getchar();cin>>year1;cin>>str2;cin>>date2;getchar();cin>>year2;int month1=the_month(str1);int month2=the_month(str2);int ans=(year2-year1)/400*97;year1%=400;year2%=400;year1+=400;year2+=400;if(year1>year2)year2+=400;for(int i=year1;i<=year2;i++)if(is_leap_year(i))ans++;if(is_leap_year(year1)&&month1>=3)ans--;if(is_leap_year(year2)){if(month2==1)ans--;if(month2==2&&date2<=28)ans--;}cout<<"Case "<<ii<<": "<<ans<<endl;}return 0;
}

求弧长

#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#define PI (2*acos(0.0))
using namespace std;
struct point
{double x,y;
};
double Distance(point a,point b)
{return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int main()
{int t;scanf("%d",&t);for(int ii=1;ii<=t;ii++){point A,B,O;double a,b,c;scanf("%lf%lf%lf%lf%lf%lf",&O.x,&O.y,&A.x,&A.y,&B.x,&B.y);a=Distance(O,A);b=Distance(O,B);c=Distance(A,B);double ans=(a+b-c)/(2*sqrt(a*b));ans=acos(ans);printf("Case %d: %lf\n",ii,ans*sqrt(a));}return 0;
}