hdu 4993(简单枚举或者hash)
2014-09-15 18:34
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4993
思路:想多了,直接枚举即可;
寻人启事:2014级新生看过来! |
Revenge of ex-EuclidTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 273 Accepted Submission(s): 162 Problem Description In arithmetic and computer programming, the extended Euclidean algorithm is an extension to the Euclidean algorithm, which computes, besides the greatest common divisor of integers a and b, the coefficients of Bézout's identity, that is integers x and y such that ax + by = gcd(a, b). ---Wikipedia Today, ex-Euclid takes revenge on you. You need to calculate how many distinct positive pairs of (x, y) such as ax + by = c for given a, b and c. Input The first line contains a single integer T, indicating the number of test cases. Each test case only contains three integers a, b and c. [Technical Specification] 1. 1 <= T <= 100 2. 1 <= a, b, c <= 1 000 000 Output For each test case, output the number of valid pairs. Sample Input 2 1 2 3 1 1 4 Sample Output 1 3 Source BestCoder Round #9 |
#include <iostream> #include <stdio.h> #include <string.h> #include <string> #include <cstdio> #include <cmath> using namespace std; int main() { int T,a,b,c; cin>>T; while(T--) { cin>>a>>b>>c; int cnt=0; for(int i=1;c-a*i>0;i++) { if((c-a*i)%b==0) cnt++; } cout<<cnt<<endl; } return 0; }PS:hash也可以做
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