蒟蒻的第24次CCF计算机软件能力认证

本文主要是介绍蒟蒻的第24次CCF计算机软件能力认证，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

100+70+40+25+0，没有技巧全是感情www!

202112-1 序列查询

这题一开始没有照题目的暗示，拿到50，后来根据题目的优化方法优化就可以了

#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <map>
#include <string>
#include <algorithm>
#include <sstream>
#include <vector>
#include <limits.h>
using namespace std;
const int maxn = 210;
int A[maxn];
int main()
{int n, N;int t = 0;long long ans = 0;int count = -1;scanf("%d %d",&n,&N);int i,j;j = 0;A[0] = 0;for (i = 1; i <= n; i++) {scanf("%d",&A[i]);}for (i = 0; i <= N-1; i++){for (; j <= n; j++){if (A[j] > i)break;}ans += j - 1;}printf("%lld\n",ans);return 0;
}

202112-2 序列查询新解

#include<bits/stdc++.h>
using namespace std;#define M 500000+2
#define LL long long
inline LL read(){LL x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
LL n,m,nn;
LL a[M];
typedef struct NODE{LL s,e;LL num;NODE(){};NODE(LL ss,LL ee,LL numm):s(ss),e(ee),num(numm){};
}node;
node g[M],f[M];
int main(){scanf("%lld%lld",&n,&nn);LL gi=0,fi=0;a[1]=read();f[fi++]=NODE(0,a[1]-1,0);for (register LL i=2;i<=n;i++){a[i]=read();f[fi++]=NODE(a[i-1],a[i]-1,i-1);}f[fi]=NODE(a[n],nn-1,n);LL r=floor(nn/(n+1));LL j=1,tmp;for (LL i=r,k=1;i<=nn-1;i+=r,k++){g[gi++]=NODE(i-r,i-1,k-1);j=i;tmp=k;}g[gi]=NODE(j,nn-1,tmp);LL ans=0,i=0,last=0;j=0;while (i<=fi&&j<=gi){LL right=min(f[i].e,g[j].e);//    cout<<right<<' ';ans+=abs(f[i].num-g[j].num)*(right-last);last=right;//    cout<<last<<endl;if (right==f[i].e) i++;if (right==g[j].e) j++;}printf("%lld\n",ans);return 0;
}

202112-3 登机牌条码

只考虑了最简单的-1的情况，用vector存放数字。

#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <map>
#include <string>
#include<cstring>
#include <algorithm>
#include <sstream>
#include <vector>
#include <limits.h>
using namespace std;
const int maxn = 1010;
vector<int> mazi;
vector<int> ans;
char a[maxn];
int main()
{int w, s;scanf("%d %d", &w, &s);scanf("%s", a);int i, j;int len = strlen(a);int flag;//flag表示当前的模式，flag=0表示大写，=1表示小写，=2表示数字flag = 0;for (i = 0; i < len; i++){if (a[i] >= 'A' && a[i] <= 'Z'){if (flag == 1)//表示之前是小写模式{mazi.push_back(28);mazi.push_back(28);flag = 0;}else if (flag == 2)//表示之前是数字模式{mazi.push_back(28);flag = 0;}mazi.push_back(a[i] - 'A');}else if (a[i] >= 'a' && a[i] <= 'z'){if (flag == 0)//大写模式{mazi.push_back(27);flag = 1;}else if (flag = 2)//数字模式{mazi.push_back(28);flag = 1;}mazi.push_back(a[i] - 'a');}else if (a[i] >= '0' && a[i] <= '9'){if (flag == 0){mazi.push_back(28);flag = 2;}else if (flag == 1)//小写{mazi.push_back(28);flag = 2;}mazi.push_back(a[i] - '0');}}if (mazi.size() % 2 == 1)//表示是偶数{mazi.push_back(29);}ans.push_back(0);for (i = 0; i < mazi.size(); i = i + 2){ans.push_back(mazi[i] * 30 + mazi[i + 1]);}if (ans.size() % w == 0){ans[0] = ans.size();}else{int temp = w - ans.size() % w;for (j = 1; j <= temp; j++){ans.push_back(900);}ans[0] = ans.size();}for (i = 0; i < ans.size(); i++){printf("%d\n", ans[i]);}return 0;
}

202112-4 磁盘文件操作

本来想用结构体，但是结构体的内存超限了，所以用了map,都用位置sta(tion)作为索引即可。

#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <map>
#include <string>
#include <algorithm>
#include <sstream>
#include <vector>
#include <limits.h>
using namespace std;
map<int, int> sta2id;//sta位置的id号
map<int, int> sta2num;//sta位置的数字
map<int, int> sta2preid;//sta位置的前id号
int main()
{int n, m, k;scanf("%d %d %d",&n,&m,&k);//读取n,m,kint i, j;//循环变量int caozuo;//操作int id,l, r, x,p;int flag;//首先初始化for (i = 1; i <= m; i++){sta2id[i] = 0;//位置对应的id号为0}for (i = 0; i < k; i++){scanf("%d",&caozuo);if (caozuo == 0){scanf("%d %d %d %d",&id,&l,&r,&x);for (j = l; j <= r; j++){if (sta2id[j] != id && sta2id[j] != 0)break;else{sta2id[j] = id;sta2num[j] = x;}                   }if (j == l)printf("-1\n");elseprintf("%d\n",j-1);}else if (caozuo == 1)//删除{scanf("%d %d %d", &id, &l, &r);flag = 0;for (j = l; j <= r; j++){if (sta2id[j] != id){flag = 1;break;}}if (flag == 1){printf("FAIL\n");}else{for (j = l; j <= r; j++){sta2id[j] = 0;//占有的id变成0；sta2preid[j] = id;//preid变为id}printf("OK\n");}}else if (caozuo == 2){scanf("%d %d %d", &id, &l, &r);flag = 0;for (j = l; j <= r; j++){if (sta2id[j] != 0 || sta2preid[j] != id){flag = 1;break;}}if (flag == 1){printf("FAIL\n");}                else{for (j = l; j <= r; j++){sta2id[j] = id;}printf("OK\n");}}else if (caozuo == 3){scanf("%d",&p);if (sta2id[p] == 0){printf("0 0\n");}else{printf("%d %d\n",sta2id[p],sta2num[p]);}}}return 0;
}