博弈专题

CodeForces - 549C

剩下的人数之和为奇数时先手胜,否则后手胜。

考虑最后一步

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main(){
 5     //ios_base::sync_with_stdio(0);
 6     int n, k;
 7     while(cin>>n>>k){
 8         int cnt = 0;
 9         int x;
10         long long sum = 0;
11         for(int i = 0;  i < n; i++){
12             scanf("%d", &x);
13             sum = sum + x;
14             if(x & 1) cnt++;
15         }
16         int ok = 1;
17         if(n == k){
18             if(sum & 1) ok = 1;
19             else ok = 0;
20         }else{
21             int m = n - k, op = m / 2;
22             if(m & 1) { // 先手最后拿
23                 if(k & 1){
24                     if(op >= cnt) ok = 0;
25                     else ok = 1;
26                 }else{
27                     if(op >= cnt || op >= (n - cnt)) ok = 0;
28                     else ok = 1;
29                 }
30             }else {
31                 if(k & 1){
32                     if(op >= (n - cnt)) ok = 1;
33                     else  ok = 0;
34                 }else{
35                     ok = 0;
36                 }
37             }
38         }
39         if(ok) cout<<"Stannis
";
40         else cout<<"Daenerys
";
41     }
42 }
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链接:here

原文地址:https://www.cnblogs.com/yijiull/p/8484474.html