As Easy As A+BTime Limit: 2000/1000 MS (Java/Others) Memory Lim
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2022-08-11 16:11:09
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小记：练习排序算法可以哦。思路：选择，冒泡，希尔，
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2023-07-18 18:38:43
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As Easy As A+B Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 53368 Accepted Submission(s): 229
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2016-05-22 23:08:00
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#include #include using namespace std;int T,a[1005],n;int main(){ scanf("%d",&T); while(T--){ scanf("%d",&n); for(int i=0;i<n;i++...
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2014-08-29 14:59:00
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思路：A、B分别对100取余的数相加再对100取余就可以了。
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2015-01-16 14:29:09
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As Easy As A+BTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/
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2022-08-05 10:46:08
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As Easy As A+BTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15920Accepted Submission(s): 6448Problem DescriptionThese days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of cou
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2011-07-28 20:37:00
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题目： As Easy As A+BTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2678 Accepted Submission(s): 1280 Problem DescriptionThese days, I am think
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2015-01-29 21:02:18
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As Easy As A+BProblem DescriptionThese days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after m
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2022-12-02 00:03:18
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As Easy A测试，输出排好序的结果Sample Input23 2 1 ...
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2023-05-24 14:46:20
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问题链接：HDU1040 As Easy As A+B。问题描述：参见上文。问题分析：这是一个排序问题，关键在于处理输入的循环控制。套路要掌握好，输入有多个case。程序说明：（略）参考链接：（略）AC的C++语言程序：/* HDU1040 As Easy As A+...
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2017-05-14 22:30:00
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As Easy As A+BTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (J
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2022-09-26 14:34:05
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最后一天差个奖励，水个A+B吧最短代码版#include <bits/stdc++.h>
int main(int a,int b)
{
return (scanf("%d%d",&a,&b),printf("%d",a+b))&0;
}某种意义上的最短代码（快读）版#include <bits/stdc++.h>
int read()
{
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2024-02-26 22:18:27
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思路：从第一个闰年开始，一直加4的
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2023-07-18 18:42:01
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Problem Description These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights. Give
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2022-09-28 11:00:15
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题目大意：给你一个数N，输出N*N的矩阵，每项为i*j。思路：两重for循环直接输出。
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2015-01-28 21:25:21
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题目链接： so easy Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 138 Accepted Submission(s): 110 Pro
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2016-03-26 23:37:00
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题目大意：给了递推公式，如果当前i%3==0，则sum(i) = sum(i-1) + i*i*i；否则sum(i) = sum(i-1) + i。思路：因为数据略大一些，所以用__int64整型来存储结果。
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2015-01-28 21:59:39
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#include<cstdio> #include<cstring> using namespace std; char s[10]; int main() { int a,b; while(scanf("%d%d",&a,&b)==2) { memset(s,0,sizeof(s)); a=a+b
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2018-03-27 13:28:00
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Problem Description These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights. Give
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2022-05-14 11:48:12
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