标签:博弈 inf 整数 out else names scan cep tle
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3598 Accepted Submission(s): 2151
Nim博弈并要求给出策略
新开两个数组记录完事
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <queue> #include <vector> #include <iomanip> #include <math.h> #include <map> using namespace std; #define FIN freopen("input.txt","r",stdin); #define FOUT freopen("output.txt","w",stdout); #define INF 0x3f3f3f3f #define INFLL 0x3f3f3f3f3f3f3f #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef long long LL; typedef pair<int, int> PII; using namespace std; int a[200005]; int p1[200005]; int p2[200005]; int main() { //FIN int n; while(~scanf("%d", &n) && n) { int ans = 0; int cnt = 0; for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); ans ^= a[i]; } for(int i = 1; i <= n; i++) { if((ans ^ a[i]) < a[i]) { p1[cnt] = i; p2[cnt] = a[i] - (ans ^ a[i]); cnt++; } } if(ans == 0) puts("No"); else { puts("Yes"); for(int i = 0; i < cnt; i++) { printf("%d %d\n", a[p1[i]], a[p1[i]] - p2[i]); } } } return 0; }
标签:博弈 inf 整数 out else names scan cep tle
原文地址:http://www.cnblogs.com/Hyouka/p/7425334.html