ybt的坑

emmmm ybt 字符串处理 例2题解错了

AC自动机板子错了（据说）

另外字符串处理的题解写的我一脸懵逼

网站上eeeee

点击查看E. 1.排队接水

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int n;
struct STU {
    int t;
    int num;
} a[1005];
bool cmp(STU a, STU b) {
    if (a.t < b.t)
        return true;
    return false;
}
long long sum[1005], tot;
int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i].t);
        a[i].num = i + 1;
    }
    sort(a, a + n, cmp);
    printf("%d", a[0].num);
    for (int i = 1; i < n; i++) {
        sum[i] = sum[i - 1] + a[i - 1].t;
        tot += sum[i];
        printf(" %d", a[i].num);
    }
    printf("\n%.2lf", tot * 1.0 / n * 1.0);
    return 0;
}
//然鹅下面就是错的
//空格格式严格
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
int n, t[10000], shunxu[10000], tem, a[100000], temp;
double pingjun, sum = 0, sum1, sum2;
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &t[i]);
        shunxu[i] = i;
        sum2 += t[i];
    }
    for (int i = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            if (t[i] > t[j]) {
                temp = t[j];
                t[j] = t[i];
                t[i] = temp;
                tem = shunxu[j];
                shunxu[j] = shunxu[i];
                shunxu[i] = tem;
            }
        }
    }
    for (int i = 1; i < n; i++) {
        printf("%d ", shunxu[i]);
    }
    printf("%d", shunxu[n]);
    for (int i = 2; i <= n; i++) {
        t[i] = t[i - 1] + t[i];
    }
    for (int i = 1; i <= n; i++) {
        sum1 += t[i];
    }
    sum = sum1 - sum2;
    pingjun = sum / n * 1.0;
    printf("\n%.2lf", pingjun);
    return 0;
}

点击查看 C.【例题3】单词替换

//90pts
#include<bits/stdc++.h>
using namespace std;
string line;
string a,b;
int main()
{
    getline(cin,line);
    cin>>a>>b;
    for(int i=0;i<line.size();i++)
    {
        int j=i;
        string word;
        while(j<line.size() && line[j]!=' ') word+=line[j++];
        i=j;
        if(word==a) cout<<b<<' ';
        else cout<<word<<' ';
    }
    return 0;
}
//100pts
#include <bits/stdc++.h>
using namespace std;
string line;
string a, b;
int main() {
    getline(cin, line);
    cin >> a >> b;
    for (int i = 0; i < line.size(); i++) {
        int j = i;
        string word;
        while (j < line.size() && line[j] != ' ') word += line[j++];
        i = j;
        if (word == a)
            cout << b << ' ';
        else
            cout << word << ' ';
    }
    return 0;
}

B. 【例题2】负环判断 :输出YE **5** N**0** 。。。face呢