Codeforces Round #827

A. Sum

找到最大值即可

#include<bits/stdc++.h>
using namespace std;

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x * f;
}

int main(){
    for( int T = read() , a , b , c ; T ; T -- ){
        a = read() , b = read() , c = read();
        if( a < b ) swap( a , b );
        if( a < c ) swap( a , c );
        cout << ( a == b + c ? "YES\n" : "NO\n" );
    }
}

B. Increasing

只要有相等的数就一定不行，所以用 set 去一下重就好

#include<bits/stdc++.h>
using namespace std;

int read(){
    int x = 0 , f = 1 , ch = getchar();
    while( (ch < '0' || ch > '9') && ch != '-' ) ch = getchar();
    if( ch == '-' ) f = -1 , ch = getchar();
    while( ch >= '0' && ch <= '9' ) x = ( x << 3 ) + ( x << 1 ) + ch - '0' , ch = getchar();
    return x * f;
}

int main(){
    for( int T = read() , n ; T ; T -- ){
        n = read();
        set<int> s;
        for( int i = 1 , x ; i <= n ; i ++ )
            x = read() , s.insert(x);
        cout << ( s.size() == n ? "YES\n" : "NO\n" );
    }
}

C. Stripes

最后一次画的一定是完整的一行或一列，并且除此之外不会有完整的行列，因为保证一行一列交替画

所有判断一下有没有完整的一行RRRRRRRR即可

#include<bits/stdc++.h>

using namespace std;

string s[10];

void solve(){
    for( int i = 0 ; i < 8 ; i ++ )
        cin >> s[i];

    for( int i = 0 ; i < 8 ; i ++ )
        if( s[i] == "RRRRRRRR" ){
            cout << "R\n";
            return;
        }

    cout << "B\n";
    return;
}

int32_t main() {
    int t ; cin >> t;
    for( ; t ; t -- )
        solve();
    return 0;
}

D. Coprime

虽然数字很多，但是数字的范围很小，我们记录一下每个数最后出现的位置，然后\(O(n^2)\)的找一下最大值就好

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int N = 1005;
int v[N] , res ;

void solve(){
    fill( v , v + N , 0 );
    res = -1;
    for( int n = read() , i = 1 , x ; i <= n ; i ++ )
        x = read() , v[x] = i;
    for( int i = 1 ; i <= 1000 ; i ++ )
        for( int j = i ; j <= 1000 ; j ++ ){
            if( v[i] == 0 || v[j] == 0 || gcd( i , j ) != 1 ) continue;
            res = max( res , v[i] + v[j] );
        }
    cout << res << "\n";
}

int32_t main() {
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

E. Scuza

求一下前缀最大值，然后在序列上二分查找就好

#include<bits/stdc++.h>
#define int long long
using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int N = 2e5+5;
int a[N] , b[N];

void solve(){
    int n = read() , q = read();
    for( int i = 1 , x ; i <= n ; i ++ )
        x = read() , a[i] = max( a[i-1] , x ) , b[i] = b[i-1] + x;
    for( int t , x ; q ; q -- ){
        x = read();
        t = upper_bound( a + 1 , a+1+n , x ) - a - 1;
        cout << b[t] << " ";
    }

    return;
}

int32_t main() {
    for( int t = read() ; t ; t -- )
        solve() , cout << "\n";
    return 0;
}

F. Smaller

首先对于s 和 t，如果希望 s小于 t，那么最优的情况就是 s 按照升序排，t 按照降序排

那么这样的话，因为一开始s 和 t 都有一个 a，所以只要 t 有非 a 的字母一定可以

如果 t 没有非 a 字母，但是 s 有则一定不可以

如果 st 都没有，s 比 t 短可以

#include<bits/stdc++.h>
#define int long long
using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int N = 2e5+5;

void solve(){
     int cntA = 1 , otherA = 0 , cntB = 1 , otherB = 0;
     string s;
     for( int q = read() , k , x ; q ; q -- ){
        k = read() , x = read() , cin >> s;
        if( k == 1 ){
            for( auto c : s ){
                if( c == 'a' ) cntA += x;
                else otherA = 1;
            }
        }
        else{
            for( auto c : s ){
                if( c == 'a' ) cntB += x;
                else otherB = 1;
            }
        }
         if( otherB )
             cout << "YES\n";
         else if( otherA == 0 && cntA < cntB )
             cout << "YES\n";
         else
             cout << "NO\n";
     }

     return;
}

int32_t main() {
    for( int t = read() ; t ; t -- )
        solve();
    return 0;
}

G. Orray

因为数字的范围只有 1e9 所以，二进制下只有 30 位，那么最多只要选出 30 个数字就好

那么我们暴力的去找哪一个数字和当前数字或起来最大的，就输出

如果所有数字或起来不会更大就不用找了，直接输出剩下所有的数字

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int N = 2e5+5;
int a[N];
bitset<N> vis;


void solve(){
    int n = read();
    vis.reset();
    for( int i = 1 ; i <= n ; i ++ )
        a[i] = read();
    for( int t = min( n , 30 ) , id , sumOR = 0 , maxOR ; t ; t -- ){
        maxOR = sumOR , id = -1;
        for( int i = 1 ; i <= n ; i ++ ){
            if( vis[i] ) continue;
            if( (sumOR | a[i]) > maxOR )
                maxOR = (sumOR | a[i]) , id = i;
        }
        if( id == -1 ) break;
        cout << a[id] << " ";
        vis[id] = 1 , sumOR |= a[id];
    }
    for( int i = 1 ; i <= n ; i ++ )
        if( !vis[i] ) cout << a[i] << " ";
    cout << "\n";
}

int main(){
    for( int t = read() ; t ; t -- )
        solve();
}