摘要: Codeforces Round #595 (Div. 3) B # include <bits/stdc++.h>using namespace std;​const int MAXN=2e5+100;int a[MAXN],b[MAXN];int dp[MAXN][2];int main(){ 阅读全文
posted @ 2022-02-26 23:01 fengzlj 阅读(48) 评论(0) 推荐(0)
摘要: Codeforces Round #601 (Div. 2) A # include <bits/stdc++.h>using namespace std;​int main(){ int T; scanf("%d",&T); while(T--){ int a,b; scanf("%d%d",&a 阅读全文
posted @ 2022-02-26 23:01 fengzlj 阅读(26) 评论(0) 推荐(0)
摘要: Poj 3437 类似于多叉树转化为二叉树 【题意】把多叉树转化为二叉树,和原来相比高度变化了多少 【题解】原来的高度就是原来最多有连续的多少条边,二叉树的高度是按照原来的树的构思每一条路径最长的那一个 # include <bits/stdc++.h> using namespace std; i 阅读全文
posted @ 2022-02-26 23:00 fengzlj 阅读(39) 评论(0) 推荐(0)
摘要: Educational Codeforces Round 74 (Rated for Div. 2) A # include<bits/stdc++.h>using namespace std;​typedef long long LL;int main(){ int t; scanf("%d",& 阅读全文
posted @ 2022-02-26 23:00 fengzlj 阅读(25) 评论(0) 推荐(0)
摘要: Codeforces Round #593 (Div. 2) A # include <bits/stdc++.h>using namespace std;​int main(){ int t; scanf("%d",&t); while(t--){ int a,b,c; scanf("%d%d%d 阅读全文
posted @ 2022-02-26 23:00 fengzlj 阅读(129) 评论(0) 推荐(0)
摘要: Codeforces Round #584 - Dasha Code Championship - Elimination Round (rated, open for everyone, Div. 1 + Div. 2) A # include <bits/stdc++.h>using names 阅读全文
posted @ 2022-02-26 22:58 fengzlj 阅读(65) 评论(0) 推荐(0)
摘要: Codeforces Round #585 (Div. 2) A # include <bits/stdc++.h>using namespace std;​int a[10],k[10];int main(){ scanf("%d%d",&a[1],&a[2]); scanf("%d%d",&k[ 阅读全文
posted @ 2022-02-26 22:58 fengzlj 阅读(13) 评论(0) 推荐(0)
摘要: Codeforces Round #590 (Div. 3) A # include <bits/stdc++.h>using namespace std;​int main(){ int q; scanf("%d",&q); while(q--){ int n; int sum=0; scanf( 阅读全文
posted @ 2022-02-26 22:58 fengzlj 阅读(33) 评论(0) 推荐(0)
摘要: Educational Codeforces Round 72 (Rated for Div. 2) A # include <bits/stdc++.h>using namespace std;​typedef long long LL;int main(){ int T; scanf("%d", 阅读全文
posted @ 2022-02-26 22:57 fengzlj 阅读(50) 评论(0) 推荐(0)
摘要: ## 2019年C++A组**E RSA解密**若d * e和(p-1) * (q-1)互质,那么说明d * e=1(mod (p-1) * (q-1) ),那么e为d的逆元,应为(p-1) * (q-1)不是指数,所以不能用费马小定理要用扩展欧几里得。求出e后,用公式X=C^e mod n就可以解 阅读全文
posted @ 2022-02-26 22:56 fengzlj 阅读(46) 评论(0) 推荐(0)