需要注意的是，要想連續搶多家銀行就得保證搶前面的銀行沒有被抓，所以樣例1中搶1和2兩家銀行的概率是 0.02 + (1-0.02)*0.03;

這裡的揹包dp[i] 中的i是val，價值；

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;

const int Maxn = 110;

double dp[10005];
int val[105]; 
double p[105];

int main (void)
{
	int T;
	cin >> T;
	for (int cas = 1; cas <= T; ++cas) {
	 	for (int i = 0; i < 10005; ++i) dp[i] = 1;
	 	int m; double limit;
		cin >> limit >> m;
		for (int i = 1; i <= m; ++i) {
			cin >> val[i] >> p[i];
		} 
		dp[0] = 0;
		for (int i = 1; i <= m; ++i) {
			for (int j = 10000-val[i]; j >= 0; --j) {
				if (dp[j] == 1) continue;
				dp[j+val[i]] = min (dp[j+val[i]], dp[j]+(1-dp[j])*p[i]);
			}
		}
		for (int i = 10000; i >= 0; --i) {
			if (dp[i] < limit || (fabs (limit-dp[i]) < 1e-6)) {
				cout << "Case " << cas << ": " << i << endl;
				break;
			}
		}
	}
	return 0;
 }