【UVA】10404-Bachet's Game（动态规划）

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如果d[i]是必胜态，那么d[i + V[j]]一定是必败态，反之亦然。

用d[i]代表棋子为i个是否为必胜态。

边界条件是d[i] = 1;

1413929110404Bachet's GameAcceptedC++0.6622014-09-03 09:44:48

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<stack>#include<queue>#include<map>#include<set>#include<list>#include<cmath>#include<string>#include<sstream>#include<ctime>using namespace std;#define _PI acos(-1.0)#define esp 1e-9typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*===========================================*/#define MAX_SIZE 1000000 + 10#define INF (1 << 20)const int MAXD = 10 + 5;int dp[MAX_SIZE];int main(){    int T;    int arr[MAXD];    while(scanf("%d",&T) != EOF){        memset(dp,0,(T + 1)*sizeof(dp[0]));        int n;        scanf("%d",&n);        for(int i = 0 ; i < n ; i++)            scanf("%d",&arr[i]);        dp[1] = 1;        for(int i = 0 ; i <= T ; i++)            for(int j = 0 ; j < n ; j++)                if(i + arr[j] <= T && !dp[i]){                     dp[i + arr[j]] = 1;                }        if(dp[T])            printf("Stan wins\n");        else            printf("Ollie wins\n");    }    return 0;}

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