Restricted RPS CodeForces

https://vjudge.net/contest/356807#problem/H

题意

　　A和B进行剪刀石头布n次

　　　　A已知B出的类型顺序，同时规定A必须出三种类型的次数

　　　　A能赢n/2局则算A赢，并给出出拳序列

　　　　否则A输

思路

　　简单模拟

　　根据给定的B出拳序列构造A出拳序列

　　不够的就填充剩余的类型

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;


bool compare(int a, int b)
{
    return a < b;//升序
}


char sequence_R[100] = {0};
char sequence_G[100] = {0};
int main(){
    int T;
    int n = 0;
    int s = 0 , r = 0 , p = 0;
    int r1 = 0 , p1 = 0 , s1 = 0;
    int sum = 0;
    cin >> T;
    while(T--)
    {
        for(int i = 0;i<n;i++)
        {
            sequence_R[i] = 0;
            sequence_G[i] = 0;
        }
        n = 0;
        s = 0; r = 0; p = 0;
        r1 = 0 , p1 = 0 , s1 = 0;
        sum = 0;
        cin >> n;
        cin >> r >> p >> s;
        cin >> sequence_R;
        for(int i = 0;i<n;i++)
        {
            if(sequence_R[i] == 'R') r1++;
            if(sequence_R[i] == 'P') p1++;
            if(sequence_R[i] == 'S') s1++;
        }
        sum += min(r1,p) + min(p1,s) + min(s1,r);
        
        if(sum >= (n+1)/2)
        {
            cout << "YES" << endl;
            for(int i = 0;i<n;i++)
            {
                if(sequence_R[i]=='R' && p > 0)
                {
                    sequence_G[i]='P';
                    p -= 1;
                } 
                else if(sequence_R[i]=='P' && s > 0)
                {
                    sequence_G[i]='S';
                    s -= 1;
                }
                else if(sequence_R[i]=='S' && r > 0) 
                {
                    sequence_G[i]='R';
                    r -= 1;
                }
            }
            for(int i = 0;i<n;i++)
            {
                if(sequence_G[i])
                {
                    cout << sequence_G[i];
                }
                else
                {
                    if(r)
                    {
                        cout << 'R';
                        r -= 1;
                    }
                    else if(p)
                    {
                        cout << 'P';
                        p -= 1;
                    }
                    else if(s)
                    {
                        cout << 'S';
                        s -= 1;
                    }
                }
            }
        }
        else
        {
            cout << "NO" << endl;
        }
        cout << endl;
    }
    return 0;
}