SGU - 403 - Scientific Problem (水)】的更多相关文章

403. Scientific Problem Time limit per test: 0.25 second(s) Memory limit: 65536 kilobytes input: standard output: standard Once upon a time Professor Idioticideasinventor was travelling by train. Watching cheerless landscape outside the window, he de…
403. Scientific Problem Time limit per test: 0.25 second(s)Memory limit: 65536 kilobytes input: standardoutput: standard Once upon a time Professor Idioticideasinventor was travelling by train. Watching cheerless landscape outside the window, he deci…
题目地址:http://acm.sgu.ru/problem.php?contest=0&problem=107 /* 题意:n位数的平方的后面几位为987654321的个数 尼玛,我看描述这一句话都看了半天,其实只要先暴力程序测试一边就知道规律 详细解释:http://www.cnblogs.com/Rinyo/archive/2012/12/04/2802089.html */ #include <cstdio> #include <iostream> #include…
题目链接: http://acm.sgu.ru/problem.php?contest=0&problem=107 题意: 平方后几位为987654321的n位数有多少个 分析: 虽然说是水题,但是我觉得很好体现了做某些数学题的方法,就是找规律 暴力求出一些较小的数,然后其他位数的数就是在求出的数的前面填数就好了. 然后注意位数很多,所以以字符的形式输出0. 代码: #include<cstdio> int main (void) { int n; scanf("%d&quo…
205. Quantization Problem time limit per test: 0.25 sec. memory limit per test: 65536 KB input: standard output: standard When entering some analog data into a computer, this information must be quantized. Quantization transforms each measured value…
题目链接: Water problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 23    Accepted Submission(s): 14 Problem Description If the numbers 1 to 5 are written out in words: one, two, three, four, fi…
一.Description Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all…
https://codeforc.es/contest/1194/problem/B 好像也没什么思维,就是一个水题,不过蛮有趣的.意思是找缺黑色最少的行列十字.用O(n)的空间预处理掉一维,然后用O(n)的时间根据另一维计算出答案. #include<bits/stdc++.h> using namespace std; typedef long long ll; int n, m; string g[50005]; int rq[50005]; int main() { #ifdef Yi…
题意:对于给定的n个字符串,可以花费a[i]  将其倒序,问是否可以将其排成从大到小的字典序,且花费最小是多少. 析:很明显的水DP,如果不是水DP,我也不会做.... 这个就要二维,d[2][maxn],d[0][i]表示第 i 个不反转是最小花费,d[1][i]表示第 i 个反转最小花费,那么剩下的就很简单了么, 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio>…
其实挺水的,因为两个数平方,只有固定的后面几位数会影响到最后结果的后面几位数.也就是说,如果想在平方之后尾数为987654321,那么就有固定的几个尾数在平方后会是这个数,打个表,发现 10^8 内 没有,10^9 内只有 8 个,然后排列组合…… 上代码: #include <cstdio> #include <cstring> #include <cstdlib> #include <cstring> using namespace std; typed…