HDU 2176 取(m堆)石子遊戲 博弈
阿新 • • 發佈:2017-08-24
ani for brush print 個數 puts tdi spa 所有
Input
輸入有多組.每組第1行是m,m<=200000. 後面m個非零正整數.m=0退出.
取(m堆)石子遊戲
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3598 Accepted Submission(s): 2151
Output 先取者負輸出No.先取者勝輸出Yes,然後輸出先取者第1次取子的所有方法.如果從有a個石子的堆中取若幹個後剩下b個後會勝就輸出a b.參看Sample Output.
Sample Input 2 45 45 3 3 6 9 5 5 7 8 9 10 0
Sample Output No Yes 9 5 Yes 8 1 9 0 10 3
Nim博弈並要求給出策略
新開兩個數組記錄完事
#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; }
HDU 2176 取(m堆)石子遊戲 博弈