先上题目: Sum Time Limit: 6000/3000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出N,a[1]... a[N],还有M,b[1]... b[M]long long ans = 0;for(int i = 1; i <= N; i ++)    for(int j = 1; j <= M; j ++)        ans +=…

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…

GCD SUM Time Limit: 8000/4000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description 给出N,M执行如下程序:long long  ans = 0,ansx = 0,ansy = 0;for(int i = 1; i <= N; i ++)   for(int j = 1; j <= M; j ++)     …

先贴代码,以后再写题解... 首先,直接枚举肯定是会超时的,毕竟n就有10^9那么多... 对于每个数,我们先把它转化为二进制:例:21-->10101: 对于00001~10101,可以分为几个部分: 00001~10000: 10001~10100: 10101 因为对于每个数,从最右边的1截断,于是就可以理解为为: 00001~10000:: 001~100: 1: 设s[i]为二进制从右边数第 i+1 个数为1 (且其他数都为0)的lowbit sum: 则 s[i]=s[i-1]*2+…

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…

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,仅仅保留最低位的1及其后面的0,截断前面的内容,然后再转成10…