hoj3433

二分加dp。dp我用的是顺推，二分查找总时间，对于每个总时间进行dp确定能否在其限制内完成任务。f[i][j]表示前i个工人完成j件a的情况下最多完成多少件b。

f[i + 1][j + k] = max(f[i + 1][j + k], f[i][j] + (t - worker[i + 1].a * k) / worker[i + 1].b);

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 55
#define maxx 1005

struct Worker
{
    int a, b;
}worker[maxn];

int n, x, y;
int f[maxn][maxx];
void input()
{
    scanf("%d%d%d", &n, &x, &y);
    for (int i = 0; i < n; i++)
        scanf("%d%d", &worker[i].a, &worker[i].b);
}

bool ok(int t)
{
    memset(f, -1, sizeof(f));
    for (int i = 0; i <= x && t - worker[0].a * i >= 0; i++)
        f[0][i] = (t - worker[0].a * i) / worker[0].b;
    for (int i = 0; i < n - 1; i++)
        for (int j = 0; j <= x; j++)
        {
            if (f[i][j] == -1)
                break;
            for (int k = 0; j + k <= x && t - worker[i + 1].a * k >= 0; k++)
                f[i + 1][j + k] = max(f[i + 1][j + k], f[i][j] + (t - worker[i + 1].a * k) / worker[i + 1].b);
        }
    if (f[n - 1][x] >= y)
        return true;
    return false;
}

int binarysearch()
{
    int l, r, mid;
    l = 0;
    r = worker[0].a * x + worker[0].b * y;
    while (l < r)
    {
        mid = (l + r) / 2;
        if (ok(mid))
            r = mid;
        else
            l = mid + 1;
    }
    return l;
}

int main()
{
//    freopen("D:\\t.txt", "r", stdin);
    int t;
    scanf("%d", &t);
    for (int i = 0; i < t; i++)
    {
        input();
        int ans = binarysearch();
        printf("Case %d: %d\n", i + 1, ans);
    }
    return 0;
}