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【UVA】10404-Bachet's Game(动态规划)

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-9
typedef 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;
}


【UVA】10404-Bachet's Game(动态规划)