CodeForces Round #553 Div2

A. Maxim and Biology

代码：

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

int N;
string s;
int minn = 0x3f3f3f3f;

int main() {
    scanf("%d", &N);
    cin >> s;
    for(int i = ; i <= N - ; i ++) {
        int cnt = ;
        for(int j = i; j < i + ; j ++) {
            if(j == i) {
                if(s[j] == 'Z') cnt += ;
                else cnt += min(s[j] - 'A', ('Z' - s[j] + ));
            }
            if(j == i + ) cnt += min(abs(s[j] - 'C'), ('Z' - s[j] + ));
            if(j == i + ) cnt += min(abs(s[j] - 'T'), (s[j] - 'A' + ));
            if(j == i + ) cnt += min(abs(s[j] - 'G'), ('Z' - s[j] + ));
        }
        minn = min(minn, cnt);
    }
    printf("%d\n", minn);
    return ;
}

/*
9
AAABBBCCC
*/

C. Problem for Nazar

代码：

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

const int mod = 1e9 + ;
long long l, r;

long long sum(long long x) {
    if(x <= ) return ;
    int flag = ;
    long long t = ;
    long long sum1 = , sum2 = ;
    for(long long i = ; i < x; ) {
        long long nx = min(i + t, x);
        if(flag) sum1 += (nx - i);
        else sum2 += (nx - i);

        t *= ;
        i = nx;
        flag ^= ;
    }
    long long ans = sum2 % mod * ((sum2 + ) % mod) % mod + sum1 % mod * (sum1 % mod) % mod;
    return ans % mod;
}

int main() {
    cin >> l >> r;
    cout << (sum(r) - sum(l - ) + mod) % mod << endl;

    return ;
}

D. Stas and the Queue at the Buffet

代码：

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

const int maxn = 1e5 + ;
int N;

struct Node {
    long long a;
    long long b;
}node[maxn];

bool cmp(const Node &n, const Node &m) {
    return (n.a - n.b) > (m.a - m.b);
}

int main() {
    scanf("%d", &N);
    for(int i = ; i <= N; i ++)
        cin >> node[i].a >> node[i].b;

    sort(node + , node +  + N, cmp);
    long long ans = ;
    for(int i = ; i <= N; i ++)
        ans += (node[i].a * (i - ) + node[i].b * (N - i));

    cout << ans << endl;

    return ;
}

E. Number of Components

代码：

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

const int maxn = 1e5 + ;
int N;
long long a[maxn];

int main() {
    scanf("%d", &N);
    for(int i = ; i <= N; i ++)
        cin >> a[i];

    long long ans = ;
    for(int i = ; i <= N; i ++) {
        if(a[i] > a[i - ])
            ans += (a[i] - a[i - ]) * (N - a[i] + );
        else if(a[i] != a[i - ])
            ans += (a[i - ] - a[i]) * a[i];
    }

    cout << ans << endl;

    return ;
}