5.[NOIP2009 普及组] 分数线

[NOIP2009 普及组] 分数线划定 题解

AC代码（C++）：

#include <bits/stdc++.h>
using namespace std;
int s[105],k[10005][105];
int main(void)
{
    int n,m,ss,kk,count = 0,i,max = 0,x,j;
    scanf("%d%d",&n,&m); // input
    m = m * 1.5; // Number of people
    for (i = 0;i < n;i++)
    {
        scanf("%d%d",&kk,&ss); // input
        k[kk][ss]++; // Included
        s[ss]++; // Included
    }
    for (i = 100;count < m;i--)
        count += s[i]; // Calculate the score and the number of people who can enter
    printf("%d %d\n",i + 1,count); // output
    x = i + 1; // Record the score line
    for(i = 100;i >= x;i--) // Enumeration from large to small within the fraction line
        for(j = 1000;j <= 10000;j++) // Search name under one score
            if(k[j][i]) printf("%d %d\n",j,i); // Search output
    return 0;
}

此处英文注释为自己靠着自己的英语知识使用百度翻译（fanyi.baidu.com/#zh/en/）所写，请酌情添加

// 解释一下
scanf("%d%d",&n,&m); // input // 输入

m = m * 1.5; // Number of people // 总人数

k[kk][ss]++; // Included // 计入

count += s[i]; // Calculate the score and the number of people who can enter // 算出分数线和能进的人数

printf("%d %d\n",i + 1,count); // output // 输出

x = i + 1; // Record the score line // 记录分数线

for(i = 100;i >= x;i--) // Enumeration from large to small within the fraction line // 分数线内从大到小枚举

for(j = 1000;j <= 10000;j++) // Search name under one score // 在一个分数下搜报名号

if(k[j][i]) printf("%d %d\n",j,i); // Search output // 搜到输出