GCD SUM Time Limit: 8000/4000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus 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 ++)       if(gcd(i,j)…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4141    Accepted Submission(s): 1441 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y)…

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7529    Accepted Submission(s): 2773 Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3315    Accepted Submission(s): 937 Problem Description   Now you get a number N, and a M-integers set, you shoul…

Co-prime Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 935    Accepted Submission(s): 339 Problem Description Given a number N, you are asked to count the number of integers between A and B in…

Description As we all know caterpillars love to eat leaves. Usually, a caterpillar sits on leaf, eats as much of it as it can (or wants), then stretches out to its full length to reach a new leaf with its front end, and finally "hops" to it by c…

题意:给定区间和n,求区间中与n互素的数的个数, . 思路:利用容斥定理求得先求得区间与n互素的数的个数,设表示区间中与n互素的数的个数, 那么区间中与n互素的数的个数等于.详细分析见求指定区间内与n互素的数的个数 容斥原理 AC代码 #include <cstdio> #include <cmath> #include <cctype> #include <bitset> #include <algorithm> #include <cs…

2301: [HAOI2011]Problem b Time Limit: 50 Sec  Memory Limit: 256 MB Submit: 7732  Solved: 3750 [Submit][Status][Discuss] Description 对于给出的n个询问,每次求有多少个数对(x,y),满足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函数为x和y的最大公约数. 100%的数据满足:1≤n≤50000,1≤a≤b≤50000,1≤c≤d≤50000…

1284 2 3 5 7的倍数 基准时间限制:1 秒 空间限制:131072 KB 分值: 5 难度:1级算法题   给出一个数N,求1至N中,有多少个数不是2 3 5 7的倍数. 例如N = 10,只有1不是2 3 5 7的倍数. Input 输入1个数N(1 <= N <= 10^18). Output 输出不是2 3 5 7的倍数的数共有多少. Input示例 10 Output示例 1经典的容斥定理,公式 |AUBUC|=|A|+|B|+|C|-|A^B|-|A^C|-|B^C|+|A…

[HDU4135]Co-prime 题意 给出三个整数N,A,B.问在区间[A,B]内,与N互质的数的个数.其中N<=10^9,A,B<=10^15. 分析 容斥定理的模板题.可以通过容斥定理求出[1,n]与x互质的数的个数.方法是先将x进行质因子分解,然后对于每个质因子pi,[1,n]内可以被pi整除的数目为n/pi.可以通过容斥定理解决逆命题,既[1,n]与x不互质的数目.n/p1+n/p2+n/p3-n/p1p2-n/p1p3-n/p2p3+n/p1p2p3.既奇数是加,偶数是减.具体的…