A. Rounding time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, V…

A. Rounding time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, V…

Codeforces Round #451 (Div. 2) A Rounding 题目链接: http://codeforces.com/contest/898/problem/A 思路: 小于等于5向下,大于补上差值输出 代码: #include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { ll n; scanf("%I64d",&n); int r=n%10;…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 模拟 [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,the main ideal 4.use the puts("") or putchar() or printf…

PROBLEM D. Alarm Clock 题 OvO http://codeforces.com/contest/898/problem/D codeforces 898d 解 从前往后枚举,放进去后不合法就拿出来,记录拿出来的次数 中途每放进去一个数,会影响到一个区间,标记这个区间的首位(做差分,首+1,尾-1),同时维护这些标记的前缀和 #include <iostream> #include <cstring> #include <cmath> #includ…

水题场.... 结果因为D题看错题意,B题手贱写残了...现场只出了A,C,E A:水题.. #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l…

B. Proper Nutrition time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative in…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 可以直接一层循环枚举. 也可以像我这样用一个数组来存y*b有哪些. 当然.感觉这样做写麻烦了.. [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,the main ideal…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 用map<string,vector > dic;模拟就好. 后缀.翻转一下就变成前缀了. 两重循环剔除这种情况不输出就好. [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 尺取法+二分. 类似滑动窗口. 即左端点为l,右端点为r. 维护a[r]-a[l]+1总是小于等于m的就好. (大于m就右移左端点) 然后看看里面的数字个数是不是小于k; 不是的话让l..r中最右边那个数字删掉就好. ->链表优化一下即可. [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself…