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…
1. sum and product rules of probability ⎧⎩⎨p(x)=∫p(x,y)dyp(x,y)=p(x|y)p(y) sum rule of probability 的积分符号自然可以换成 ∑ 求和符号(针对离散型随机变量) 2. 简单应用 sum and product rules of probability in Bishop's book sum and product rules of probability 证明:p(x=1|D)=∫10p(x=1|μ…
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…
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…
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…
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers…
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers…
