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C 贪心 写的时候突然发现这么容易，所有的绳子都要拆掉，而且绳子的个数固定，所以只要每次拆绳子，只要找绳子两端v小的即可，O(n) //代码里面有没用的冗余
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;const int MAXN = 2000*2+100;
vector<int>g[MAXN];
int v[MAXN],id[MAXN];
int n,m;int main()
{//IN("C.txt");while(~scanf("%d%d",&n,&m)){int u,t;for(int i=0;i<=n;i++)g[i].clear();for(int i=1;i<=n;i++)scanf("%d",&v[i]),id[i]=i;//sort(id+1,id)ll ans=0;for(int i=0;i<m;i++){scanf("%d%d",&u,&t);g[u].push_back(t);g[t].push_back(u);if(v[t]>v[u])ans+=v[u];else ans+=v[t];}printf("%I64d\n",ans);}return 0;
}
B---胡蒙的，至今不解为啥按lowbit从大到小一定可以找出sum的组合
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;const int MAXN = 1e5+100;int vis[MAXN],a[MAXN];
int sum,up;inline int lowbit(int x)
{return x&(-x);
}vector<int>ans;bool cmp(int ca,int cb)
{return lowbit(ca)>lowbit(cb);
}int main()
{//IN("B.txt");while(~scanf("%d%d",&sum,&up)){ans.clear();for(int i=0;i<=up;i++)a[i]=i;sort(a+1,a+1+up,cmp);for(int i=1;i<=up;i++){int tmp=lowbit(a[i]);// printf("i=%d %d\n",i,a[i]);/if(sum>=tmp)sum-=tmp,ans.push_back(a[i]);if(sum==0)break;}if(sum!=0)puts("-1");else{printf("%d\n",ans.size());if(ans.size())printf("%d",ans[0]);for(int i=1;i<ans.size();i++)printf(" %d",ans[i]);putchar('\n');}}return 0;
}
A 纯属联系string类的substr函数了 不过用String数组写更好些，我的代码冗余严重
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;string a,b,c,d;int checka()
{int len=a.size()*2;if(len <= b.size() && len <=c.size() && len<=d.size())return 1;if(a.size()>=b.size()*2 && a.size()>=d.size()*2 && a.size()>=c.size()*2)return 1;return 0;
}
int checkb()
{int len=b.size()*2;if(len <= a.size() && len <=c.size() && len<=d.size())return 1;if(b.size()>=a.size()*2 && b.size()>=d.size()*2 && b.size()>=c.size()*2)return 1;return 0;
}int checkc()
{int len=c.size()*2;if(len <= a.size() && len <=c.size() && len<=d.size())return 1;if(c.size()>=a.size()*2 && c.size()>=d.size()*2 && c.size()>=b.size()*2)return 1;return 0;
}int checkd()
{int len=d.size()*2;if(len <= a.size() && len <=c.size() && len<=b.size())return 1;if(d.size()>=a.size()*2 && d.size()>=c.size()*2 && d.size()>=b.size()*2)return 1;return 0;
}int main()
{//IN("A.txt");while(cin >> a >> b >> c >> d){a=a.substr(2,a.size()-2);b=b.substr(2,b.size()-2);c=c.substr(2,c.size()-2);d=d.substr(2,d.size()-2);int flag=0;char ans;if(checka()){ans='A';flag++;}if(checkb()){ans='B';flag++;}if(checkc()){ans='C';flag++;}if(checkd()){ans='D';flag++;}if(flag == 1){printf("%c\n",ans);continue;}puts("C");}return 0;
}
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