Codeforces Round #784 (Div. 4), problem: (C) Odd/Even Increments

直接将flag写到循环中，简化代码量

前一种解决方案

#include<iostream> 
#include<cstring>
using namespace std;

const int N = 2e5+5;

int a[N];


int main(){
    int T;
    cin >> T;

    while(T--){
        memset(a,0,sizeof a);
        int n;
        cin >> n;

        for(int i = 1; i <= n; i++){
            cin >> a[i]; 
        }

        int flag = a[1]%2,wa = 0;;
        for(int i = 3; i <= n; i+=2){
            if(a[i]%2 != flag) wa = 1;
        }

        if(wa){
            puts("NO"); 
            continue;
        } 

        flag = a[2]%2,wa = 0;;
        for(int i = 4; i <= n; i += 2){
            if(a[i]%2 != flag) wa = 1;
        }

        if(wa){
            puts("NO");
        }
        else{
            puts("YES");
        }
    }
}

后一种解决方案

#include<iostream> 
#include<cstring>
using namespace std;

const int N = 110;

int a[N];

int main(){
    int T;
    cin >> T;

    while(T--){ 
        int n;
        cin >> n;

//      memset(a,0,sizeof a);

        for(int i = 1; i <= n; i++){
            cin >> a[i]; 
        }

        bool ok = true;
        for(int i = 3; i <= n && ok; i += 2){
            if(a[i]%2 != a[1]%2) ok = false;
        }

        for(int i = 4; i <= n && ok; i += 2){
            if(a[i]%2 != a[2]%2) ok = false;
        }

        if(!ok){
            puts("NO");
        }
        else{
            puts("YES");
        }
    }
    return 0;
}

Codeforces Round #784 (Div. 4), problem: (D) Colorful Stamp

使用或运算来表示两种变量，避免对初始情况的讨论

我的解决方案：

 #include<iostream> 
#include<cstring>
using namespace std;

const int N = 2e5+5;

int a[N];


int main(){
    int T;
    cin >> T;

    while(T--){
        memset(a,0,sizeof a);
        int n;
        cin >> n;

        string s;
        cin >> s;

        int flag = 1,b_flag = 0,r_flag = 0;
        for(int i = 0; i <= n; i++){
            if((s[i] == 'W' && i != 0) || i == n){
                if(b_flag == 0 && r_flag == 0) continue;
                if(b_flag == 0 || r_flag == 0) flag = 0;
                b_flag = 0,r_flag = 0;  
            } 
            if(s[i] == 'B') b_flag = 1;
            if(s[i] == 'R') r_flag = 1;
        }

        if(flag) puts("YES");
        else puts("NO");
    }

    return 0;
}

更简洁的解决方案：

 #include<iostream> 
#include<cstring>
using namespace std;

const int N = 2e5+5;

int a[N];


int main(){
    int T;
    cin >> T;

    while(T--){
        memset(a,0,sizeof a);
        int n;
        cin >> n;

        string str;
        cin >> str;

        int s = 0,ok = 1;
        for(int i = 0; i <= n && ok; i++){
            if(str[i] == 'W' || i == n){
                if(s == 1 || s == 2) ok = 0;
                s = 0;
            } 
            if(str[i] == 'B') s |= 1;
            if(str[i] == 'R') s |= 2;
        }

        if(ok) puts("YES");
        else puts("NO");
    }

    return 0;
}

还有一点就是表达两种条件只能有一种成立
(表达式1)+(表达式2) == 1