hdu 2955 Robberies dp

题意：Roy抢银行，n个银行中每个银行有val【i】的钱，Roy在这个银行被抓的概率为p【i】，而Roy有一个安全线P，当Roy抢银行的安全程度在P以内使安全的，否则就会被抓。问Roy不被抓时最多能抢多少钱

//设dp【i】表示安全情况下偷i钱 的最大安全程度。在P以内是安全的，则在1-P要被抓，所以有dp【n】>=1-P。p【i】表示被抓的概率，则1-p【i】为安全的

那么可以得到dp[j] = max( dp[j], dp[j-v[i]] * p[i] );

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <time.h>
#include <vector>
#include <cstdio>
#include <string>
#include <iomanip>
///cout << fixed << setprecision(13) << (double) x << endl;
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define Mp(a, b) make_pair(a, b)
#define asd puts("asdasdasdasdasdf");
typedef long long ll;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int N = 200100;
const ll mod = 1e9+7;

double P;
double dp[N];
int n;
double p[N];
int v[N], w[N];

int main()
{
	int tot;
	for( scanf("%d", &tot); tot--; ) {
		scanf("%lf %d\n", &P, &n);
		P = 1-P;
		int sum = 0;
		for( int i = 1; i <= n; ++i ) {
			scanf("%d%lf", &v[i], &p[i]);
			p[i] = 1 - p[i];
			sum += v[i];
		}
		for( int i = 1; i <= sum; ++i )
			dp[i] = 0;
		dp[0] = 1;
		for( int i = 1; i <= n; ++i ) {
			for( int j = sum; j >= v[i]; --j ) {
				dp[j] = max( dp[j], dp[j-v[i]] * p[i] );
			}
		}
		for( int i = sum; i >= 0; --i ) {
			if( dp[i] - P >= eps ) {
				printf("%d\n", i);
				break;
			}
		}
	}
	return 0;
}