Codeforces Edu Round 56 A-D

A. Dice Rolling

把\(x\)分解为\(a * 6 + b\)，其中\(a\)是满6数，\(b\)满足\(1 <= b < 6\)，即可...

#include <iostream>
#include <cstdio>
using namespace std;
int main(){
    int T; scanf("%d", &T);
    while(T--){
        int x; scanf("%d", &x);
        printf("%d\n", x % 6 ? x / 6 + 1 : x / 6);
    }

    return 0;
}

B. Letters Rearranging

判断，如果回文就直接调整一位即可。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1010;
char s[N];
int cnt[26], tot = 0, loc = 0;
int main(){
    int T; scanf("%d", &T);
    while(T--){
        memset(cnt, 0, sizeof cnt); loc = tot = 0;
        bool ep = false;
        scanf("%s", s + 1);
        int n = strlen(s + 1);
        for(int i = 1; i <= n; i++){
            cnt[s[i] - 'a'] ++;
        }
        for(int i = 0; i < 26; i++){
            if(cnt[i]) tot++, loc = i;
        }
        if(tot == 1) puts("-1");
        else{
            for(int i = 1; i <= n; i++)
                if(s[i] == s[n - i + 1] && i != n - i + 1){
                    for(int j = 1; j <= n; j++){
                        if(i != j && j != n - i + 1){
                            if(s[j] != s[i]){
                                swap(s[j], s[i]);
                                printf("%s\n", s + 1);
                                ep = true; break;
                            }
                        }
                    }
                    if(ep) break;
                }
            if(!ep) printf("%s\n", s + 1);
        }
    }
    return 0;
}

C. Mishka and the Last Exam

贪心构造，尽量使每个\(a[i] (1 <= i <= n / 2)\)最小：

在符合\(a[i - 1] <= a[i]\)的条件下，也要满足\(a[n - i + 1] <= a[n - i + 1 + 1]\)，所以说这个界限为：

\(a[i] = max(a[i - 1], b[i] - a[n - i + 1 + 1])\) 注意边界，把\(a[n + 1]\) 赋值为极大值。

#include <iostream>
#include <cstdio>
#include <limits.h>
using namespace std;
typedef long long LL;
const int N = 200010;
int n;
LL a[N], b[N >> 1];
int main(){
    scanf("%d", &n);
    a[n + 1] = 9223372036854775807ll;
    for(int i = 1; i <= (n >> 1); i++) {
        scanf("%lld", b + i);
        a[i] = max(a[i - 1], b[i] - a[n - i + 1 + 1]);
        a[n - i + 1] = b[i] - a[i];
    }
    for(int i = 1; i <= n; i++)
        printf("%lld ", a[i]);
    return 0;
}

D. Beautiful Graph

实质是一个二分图染色问题，对于选定每个奇数后，偶数是对应主线的，所以只需算奇数的方案即可。

对于每一个联通快而言相互没有影响，它们的方案是：

\(2 ^ {cnt0} + 2 ^ {cnt1}\)其中两个\(cnt\)代表二分图染色两个色彩的数量，颜色可以调换，且填奇数，每个位置有两种选择...

答案就是每个联通快的相乘，注意如果二分图失败了就\(ans = 0\)吧

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
typedef long long LL;
const int N = 300010, M = 300010, MOD = 998244353;
int n, m, numE, head[N], f[N], ans;
struct Edge{
    int next, to;
}e[M << 1];
void addEdge(int from, int to){
    e[++numE].next = head[from];
    e[numE].to = to;
    head[from] = numE;
}
queue<int> q;
int power(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = (LL)res * a % MOD;
        a = (LL)a * a % MOD;
        b >>= 1;
    }
    return res;
}
int main(){
    int T; scanf("%d", &T);
    while(T--){
        while(q.size()) q.pop();
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) head[i] = 0, f[i] = -1;
        numE = 0; ans = 1;
        for(int i = 1; i <= m; i++){
            int u, v; scanf("%d%d", &u, &v);
            addEdge(u, v); addEdge(v, u);
        }
        for(int i = 1; i <= n; i++){
            if(~f[i]) continue;
            int cnt0 = 1, cnt1 = 0;
            f[i] = 0; q.push(i);
            while(!q.empty()){
                int u = q.front(); q.pop();
                for(int i = head[u]; i; i = e[i].next){
                    int v = e[i].to;
                    if(f[v] == -1){
                        f[v] = f[u] ^ 1;
                        if(!f[v]) cnt0++;
                        else cnt1++;
                        q.push(v);
                    }else if(f[v] != (f[u] ^ 1)){
                        ans = 0; break;
                    }
                }
                if(!ans) break;
            }
            if(!ans) break;
            ans = (ans * ((LL)(power(2, cnt0) + power(2, cnt1)) % MOD)) % MOD;
        }
        printf("%d\n", ans);
    }
    return 0;
}