noip模拟赛 楼

分析：题目可以转化为对于一个数，对它进行x次减法操作，n-x次加法操作，使他变成最小的非负整数.因为每减一次数就会减小，次数是一定的，所以可以二分x,就可以了.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

const int inf = 0x7fffffff;

int n, m, u[], v[];
ll ans = inf;

void erfen(int cur)
{
    int l = , r = ,res = ;
    while (l <= r)
    {
        int mid = (l + r) >> ;
        if (1LL * u[cur] * (n - mid) - 1LL * v[cur] * mid >= )
        {
            res = mid;
            l = mid + ;
        }
        else
            r = mid - ;
    }
    ans = min(ans, 1LL * u[cur] * (n - res) - 1LL * v[cur] * res);
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = ; i <= m; i++)
        scanf("%d%d", &u[i], &v[i]);
    for (int i = ; i <= m; i++)
        erfen(i);
    printf("%lld\n", ans);

    return ;
}