hdu2955

给定一个被抓的概率p，还有一些银行，给定这些银行的钱数以及抢劫这些银行被抓的概率。

求在被抓的概率不大于p的情况下最多可以抢到多少钱。

01背包。

dp：
dp[i]表示抢劫i元逃跑的概率，则dp[i] = max(, dp[i], dp[i - v[j]] * (1 - w[i]));

抢劫多个银行逃跑的概率等于抢劫每个银行逃跑概率之积。

#include <iostream>#include <cmath>#include <algorithm>#include <string>#include <deque>#include <cstring>#include <cstdio>#include <vector>#include <set>#include <map>#include <queue>#include <cstdlib>#include <iomanip>using namespace std;#define rep(i, a, b) for( i = (a); i <= (b); i++)#define reps(i, a, b) for( i = (a); i < (b); i++)#define pb push_back#define ps push#define mp make_pair#define CLR(x,t) memset(x,t,sizeof x)#define LEN(X) strlen(X)#define F first#define S second#define Debug(x) cout<<#x<<"="<<x<<endl;const double euler_r = 0.57721566490153286060651209;const double pi = 3.141592653589793238462643383279;const double E = 2.7182818284590452353602874713526;const int inf=~0U>>1;const int MOD = int(1e9) + 7;const double EPS=1e-6;typedef long long LL;int t, n, i, j, ans, v[105], sum;double w[105], p, f[10005];int main(){    cin >> t;    while(t--)    {        cin >> p >> n;        sum = 0;        rep(i, 1, n) { cin >> v[i] >> w[i]; sum += v[i];}        //CLR(f, -1);        rep(i, 1, 10004) f[i] = -1;        f[0] = 1;//抢钱的逃跑的概率        rep(i, 1, n)        {            for(j = sum; j >= v[i]; j--)            {                f[j] = max(f[j], f[j - v[i]] * (1 - w[i]));            }        }        for(j = sum; j >= 0; j--)            if(f[j] >= 1 - p)            {                ans = j;                break;            }        cout << ans << endl;    }    return 0;}