简单思维题 #include<bits/stdc++.h> using namespace std; #define int long long #define inf 0x3f3f3f3f3f3f3f ]; signed main(){ int n;cin>>n; ; ;i<=n;i++) { cin>>a[i]; if(a[i]<minx){ x=i; minx=min(minx,a[i]); } } ){ cout<<"-1&qu…
codeforces 535A-水题: #include <bits/stdc++.h> using namespace std; typedef long long LL; char s2[15][20]={"eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen",&q…
题意:给一个n<=1e7,找两个合数a和b使得a-b的差为n. 构造a=3n,b=2n,必含有公因子n,只有当n是1的时候是特例. #include<bits/stdc++.h> using namespace std; #define int long long #define inf 0x3f3f3f3f3f3f #define N 300009 ,,,,}; signed main(){ int n;scanf("%lld",&n); ){ cout&l…
//在我对着D题发呆的时候,柴神秒掉了D题并说:这个D感觉比C题简单呀!,,我:[哭.jpg](逃 Codeforces Round #435 (Div. 2) codeforces 862 A. Mahmoud and Ehab and the MEX[水] 题意:定义一个集合的MEX为其中的最小的不在集合里的非负整数,现在给一个大小为N的集合和X,每次操作可以向其中加入一个非负整数或删除一个数,问最小的操作次数,使得这个集合的MEX为X. #include<cstdio> #include…
Codeforces Round #441 (Div. 2) codeforces 876 A. Trip For Meal(水题) 题意:R.O.E三点互连,给出任意两点间距离,你在R点,每次只能去相邻点,要走过n个点,求走过的最短距离. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { int n, a, b, c; scanf("…
Codeforces Round #440 (Div. 2) codeforces 870 A. Search for Pretty Integers(水题) 题意:给两个数组,求一个最小的数包含两个数组各至少一个数. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n, m; ], b[]; int main() { , y = , s = ; scanf(…
A. The Useless Toy time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which…
A. Saitama Destroys Hotel time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an eleva…
A. Joysticks time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at…
比赛链接:https://codeforces.com/contest/1433 A. Boring Apartments 题解 模拟即可. 代码 #include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { int n; cin >> n; int ans =…
CF比赛题目地址:http://codeforces.com/contest/862 A. Mahmoud and Ehab and the MEX ·英文题,述大意: 输入n,x(n,x<=100),然后输入n个数作为一个集合.现在可以进行多次操作,每次操作为删除一个数或者添加一个数,求最少操作数使得集合Mex值等于x. ·分析: 也许是考察Mex的吧~那就当复习了,我们的策略是,如果当前集合的 Mex比x小,那就一直添加数;如果Mex比x大,那么就删除那个和x相同的元素…
B. Inventory time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inv…
#include<bits/stdc++.h> using namespace std; #define inf 0x3f3f3f3f3f3f #define int long long ]; signed main(){ int n,k; cin>>n>>k; string str; cin>>str; map<int,int> mp; ; ;i<str.size();i++){ if(!mp[str[i]-'A']){ mp[str[i…
A题 题意:给你 n 个数 , 你需要改变这些数使得这 n 个数的值相等 , 并且要求改变后所有数的和需大于等于原来的所有数字的和 , 然后输出满足题意且改变后最小的数值. AC代码: #include<bits/stdc++.h> using namespace std; #define int long long signed main(){ int _; cin>>_; while(_--){ int n; cin>>n; ]; ; ;i<=n;i++){…
A题 题意:给你 n 个数 , 你需要改变这些数使得这 n 个数的值相等 , 并且要求改变后所有数的和需大于等于原来的所有数字的和 , 然后输出满足题意且改变后最小的数值. AC代码: #include<bits/stdc++.h> using namespace std; #define int long long signed main(){ int _; cin>>_; while(_--){ int n; cin>>n; ]; ; ;i<=n;i++){…
比赛链接:https://codeforces.com/contest/1440 A. Buy the String 题解 枚举字符串中 \(0\) 或 \(1\) 的个数即可. 代码 #include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { int n, c0, c1…
比赛链接:https://codeforces.com/contest/1438 A. Specific Tastes of Andre 题意 构造一个任意连续子数组元素之和为子数组长度倍数的数组. 题解 构造全为同一值的任意数组即可. 代码 #include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t…
比赛链接:https://codeforces.com/contest/1436 A. Reorder 题解 模拟一下这个二重循环发现每个位置数最终都只加了一次. 代码 #include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; cin >> t; while (t--) { int n, m; cin >>…
