Codeforces Round #553 (Div. 2)

A. Maxim and Biology

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int route[26][26];
void solve() {
    for(int u = 0; u < 26; u += 1) {
        for(int v = 0; v < 26; v += 1) {
            if(u <= v)
                route[u][v] = min(abs(v - u), u + 26 - v);
            else
                route[u][v] = min(abs(u - v), v + 26 - u);
        }
    }
}
int getLength(string s, int i) {
    return route[s[i] - 'A'][0] + route[s[i + 1] - 'A']['C' - 'A'] + route[s[i + 2] - 'A']['T' - 'A'] + route[s[i + 3] - 'A']['G' - 'A'];
}
int main() {
    solve();
    int n;
    string s;
    cin >> n >> s;
    int length = s.size();
    int Min = 100;
    for(int i = 0; i < length - 3; i += 1) {
        Min = min(Min, getLength(s, i));
    }
    cout << Min;
    return 0;
}

B. Dima and a Bad XOR

暴力

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int N, M, x, vis[1024], result, G[505][505];
bool sign;
vector<int> vec[505], path, lastpath;
void DFS(int dep, int n) {
    if(sign)
        return ;
    if(dep == N + 1) {
        if(n != 0) {
            result = n;
            lastpath = path;
            sign = true;
        }
        return ;
    }
    int length = vec[dep].size();
    for(int i = 0; i < length; i += 1) {
        int T = vec[dep][i];
        path.push_back(T);
        DFS(dep + 1, n ^ G[dep][T]);
        path.pop_back();
    }
    return ;
}
int main() {
    scanf("%d%d", &N, &M);
    for(int r = 1; r <= N; r += 1) {
        memset(vis, 0, sizeof(vis));
        for(int c = 1; c <= M; c += 1) {
            scanf("%d", &G[r][c]);
            if(vis[G[r][c]])
                continue;
            vis[G[r][c]] = 1;
            vec[r].push_back(c);
        }
    }
    sign = false;
    result = 0;
    DFS(1, 0);
    if(result == 0)
        printf("NIE");
    else {
        printf("TAK\n");
        for(int i = 0; i < lastpath.size(); i += 1) {
            if(i != 0)
                printf(" ");
            printf("%d", lastpath[i]);
        }
    }
    return 0;
}

C. Problem for Nazar

题意：有两个集合，一个是奇数集合(1,3,5,7,…)，一个是偶数集合 (2,4,6,8,…)，现构造如下序列，第一次操作，从奇数集合中取1个数(1)，第二次操作，从偶数序列中取2两个数(2,4)，第三次操作从奇数集合中取4个数（3,5,7,9），第四次操作从偶数集合中取8个数(6,8,10..)以此类推，构造一个无穷大的数列，题目给你一个区间[l,r]问，区间和是多少？

思路：求区间和，那么想到一个求区间和的公式S[l..r]=S[1...r]-S[1...l-1]，分别统计区间[1...r]和[1...l-1]当中奇数和偶数的个数，等差数列求和然后作差

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int M = 1e9 + 7;
const int N = 1e2;
long long solve(long long x) {
    int k = 1;
    long long n[2];
    memset(n, 0, sizeof(n));
    long long s = 1;
    while(x) {
        n[k] += min(s, x);
        x -= min(x, s);
        s <<= 1;
        k ^= 1;
    }
    for(int i = 0; i < 2; i += 1)
        n[i] = n[i] % M;
    long long res = (2 * n[0] % M + n[0] * max(0LL, n[0] - 1) % M) % M;
    res = (res + n[1] % M + n[1] * max(0LL, n[1] - 1) % M) % M;
    return res;
}
int main() {
    lo  ng long l, r;
    scanf("%lld%lld", &l, &r);
    printf("%lld\n", (solve(r) - solve(l - 1) + M) % M);
    return 0;
}

D. Stas and the Queue at the Buffet

题意：初始有n个元素，每个元素有两种属性(a,b)，每个元素对序列的贡献的计算公式是a⋅(j−1)+b⋅(n−j) ，j的含义是该元素在序列中的位置，现在希望你能找到一种排列，使得所有元素对序列的贡献和最小，输出这个最小值

思路：计算元素贡献的式子可以拆开 a⋅(j−1)+b⋅(n−j) = (a-b)j+bn-a,我们发现bn-a与元素所处的位置无关，所以现在需要我们讨论的问题就是使前半部分(a-b)j尽量小，a-b已知，怎么安排j也很显然
 

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 1e5 + 5;
long long a[N], b[N], c[N], res;
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i += 1) {
        scanf("%lld%lld", &a[i], &b[i]);
        c[i] = a[i] - b[i];
        res += b[i] * n - a[i];
    }
    sort(c + 1, c + 1 + n);
    for(int i = 1; i <= n; i += 1) {
        res += c[i] * (n - i + 1);
    }
    printf("%lld\n", res);
    return 0;
}

E. Number of Components

官方题解:

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 1e5 + 5;
int a[N];
long long res;
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i += 1)
        scanf("%d", &a[i]);
    for(int i = 1; i <= n; i += 1) {
        if(a[i] > a[i - 1])
            res += (long long)(a[i] - a[i - 1]) * (n - a[i] + 1); //l的取值个数为(a[i] - a[i - 1]),r的取值个数为(n - a[i] + 1)
        if(a[i] < a[i - 1])
            res += (long long)(a[i - 1] - a[i]) * a[i]; //l的取值个数为a[i]，r的取值个数为(a[i - 1] - a[i])
    }
    printf("%lld\n", res);
    return 0;
}