Divide Sum Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description long long ans = 0;for(int i = 1; i <= n; i ++)    for(int j = 1; j <= n; j ++)        ans += a[i] / a[j];给出n,a…

Divide Sum Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description long long ans = 0;for(int i = 1; i <= n; i ++)    for(int j = 1; j <= n; j ++)        ans += a[i] / a[j];给出n,a[1]...a[n],求an…

Sum Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description You are given an N*N digit matrix and you can get several horizontal or vertical digit strings from any position. For…

Sum vs Product Time Limit: 4000/2000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description Peter has just learned mathematics. He learned how to add, and how to multiply. The fact that 2 + 2 = 2 × 2…

Sum vs Product Time Limit: 2000/1000MS (Java/Others)    Memory Limit: 128000/64000KB (Java/Others) Submit Status Problem Description Peter has just learned mathematics. He learned how to add, and how to multiply. The fact that 2 + 2 = 2 × 2 has amaze…

先上题目: Power Sum Time Limit: 20000/10000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出n,m,p,求 (1^m + 2^m + 3^m + 4^m + ... + n^m) % p Input 第一行一个数T( <= 10),表示数据总数 然后每行给出3个数n,m,p(1 <= n <= m <= 10…

先上题目: C - Lowbit Sum Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description long long ans = 0;for(int i = 1; i <= n; i ++)    ans += lowbit(i)lowbit(i)的意思是将i转化成二进制数之后,只保留最低位的1及其后面的0,截断前面的内容,然…

解题思路:ACM紫书 第十章 P319 有重复元素的全排列 答案: 所有数的和的阶乘 除以 每个数阶乘的乘积 因为给定 (26*12)! 会爆掉(long long),这里用java 的BigInteger. import java.math.BigInteger; import java.util.Scanner; public class Main { public static BigInteger jc(BigInteger n) { if (n.compareTo(new BigInt…

Curry: The idea of Curry is to spreate the data from the function. Using Curry to define the function logic and later pass the data into the function logic. Example1: const get = R.curry(function(prop, obj){ return obj[prop]; }); const obj1 = { foo:…

增长率 = (DIVIDE(SUM('业绩达成'[实际业绩]),CALCULATE(SUM('业绩达成'[实际业绩]),PREVIOUSMONTH('业绩达成'[周期])))-1)*100上月业绩 = CALCULATE(SUM('业绩达成'[实际业绩]),PREVIOUSMONTH('业绩达成'[周期]))…