Codeforces 1221E Game With String（string分a，b变换，博弈）

链接：http://codeforces.com/contest/1221/problem/E

题意：给出有.x组成的字符串，两个人博弈，一个人每次可以选择一段连续长度为a的.使得它变成x，另一个选择长度为b的进行操作，为谁会输，保证a>b

题解：

    #include <bits/stdc++.h>
    using namespace std;
     
    const int maxn=3e5+5;
    char s[maxn];
    int T, n, x, y, a[maxn];
    vector<int> v;
     
    int main(){
       // freopen("in.txt", "r", stdin);
        for(cin>>T; T--; )
        {
            cin>>x>>y;
            cin>>(s+1);
            n=strlen(s+1);
            for(int i=1; i<=n; i++)
                a[i]= s[i]=='.'?1:0;
            int cnt=0, ok=0;
            v.clear();
            for(int i=1; i<=n; i++){
                if(a[i]) ++cnt;
                else if(cnt) v.push_back(cnt), cnt=0;
            }
            if(cnt) v.push_back(cnt), cnt=0;
            for(auto &val: v){
                if(val<x&&val>=y) ok=1;
                if(val>=2*y) ++cnt;
            }
     
            if(ok || cnt>=2){
                cout<<"No"<<"
"; continue;
            }
            if(cnt==0){
                for(auto &val: v) if(val>=x) ++cnt;
                if(cnt&1){
                    cout<<"Yes"<<"
"; continue;
                }else{
                    cout<<"No"<<"
"; continue;
                }
            }
            if(cnt==1){
                int max_len=*max_element(v.begin(), v.end());
                int flag=0; cnt=0;
                for(auto &val: v) if(val!=max_len&&val>=x) ++cnt;
                for(int l=1; l+x-1<=max_len; l++)
                {
                    int r=l+x-1;
                    int left=l-1, right=max_len-r;
                    if(left>=2*y || right>=2*y) continue;
                    if((left>=y&&left<x) || (right>=y&&right<x)) continue;
                    int tmp=cnt+(left>=x)+(right>=x);
                    if(tmp%2==0){
                        flag=1; cout<<"Yes"<<"
"; break;
                    }
                }
                if(!flag){
                    cout<<"No"<<"
"; continue;
                }
            }
     
        }
        return 0;
    }