ural 1869. New Year Cruise - c++编程基础

描述：从Vladivostok站Moscow站共有n个站台，Vladivostok记为1，Moscow记为n，火车来回往返。现在给出从车站i到车站j和j到i的买票人数，计算需要的最少节车厢。

思路：水题，模拟即可。注意先下后上，用一个max保持最大的人数。

代码：

//g++ 4.7.2  用到了变长数组  
#include   
int main()  
{  
    int n;  
    scanf("%d", &n);  
    int a[n+1][n+1];  
    for (int i = 1; i <= n; ++i)  
        for (int j = 1; j <= n; ++j)  
            scanf("%d", &a[i][j]);  
    int max = 0, sum = 0;  
    for (int i = 1; i <= n; ++i)  
    {  
        for (int t = 1; t < i; ++t)  
            sum -= a[t][i];  
        for (int j = i + 1; j <= n; ++j)  
        {  
            sum += a[i][j];  
            if (sum >




 max)  
                max = sum;  
        }  
    }  
    sum = 0;  
    for (int i = n; i >= 1; --i)  
    {  
        for (int t = n; t > i; --t)  
            sum -= a[t][i];  
        for (int j = i - 1; j >= 1; --j)  
        {  
            sum += a[i][j];  
            if (sum > max)  
                max = sum;  
        }  
    }  
    printf("%d\n", max % 36 == 0   max / 36 : max / 36 + 1);  
    return 0;  
}