PAT 天梯赛 L2-009. 抢红包 【排序】

题目链接

https://www.patest.cn/contests/gplt/L2-009

思路

用结构体存储,然后结构体排序 注意一下 个人编号是从 1 开始
计数的
AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>

using namespace std;
typedef long long LL;

const double PI  = 3.14159265358979323846264338327;
const double E   = 2.718281828459;
const double eps = 1e-6;

const int MAXN = 0x3f3f3f3f;
const int MINN = 0xc0c0c0c0;
const int maxn = 1e4 + 5;
const int MOD  = 1e9 + 7;

struct Node
{
    int tot, id;
    double sum;
}q[maxn];

bool comp(Node x, Node y)
{
    if (x.sum == y.sum)
    {
        if (x.tot == y.tot)
            return x.id < y.id;
        return x.tot > y.tot;
    }
    return x.sum > y.sum;
}

int main()
{
    memset(q, 0, sizeof(q));
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        q[i].id = i;
    for (int i = 0; i < n; i++)
    {
        int k;
        scanf("%d", &k);
        int id, num;
        for (int j = 0; j < k; j++)
        {
            scanf("%d%d", &id, &num);
            q[id - 1].sum += num;
            q[id - 1].tot ++;
            q[i].sum -= num;
        }
    }
    sort (q, q + n, comp);
    for (int i = 0; i < n; i++)
        printf("%d %.2lf
", q[i].id + 1, q[i].sum / 100);
}
原文地址:https://www.cnblogs.com/Dup4/p/9433263.html