HDU 4993 Revenge of ex-Euclid(数学题 暴力)
2014-09-13 18:28
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4993
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
Sample Output
Source
BestCoder Round #9
题意:
寻找ax+by = c有多少组解!
代码如下:
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
题意:
寻找ax+by = c有多少组解!
代码如下:
#include <cstdio> #include <cmath> int main() { int t; int a,b,c; scanf("%d",&t); while(t--) { scanf("%d%d%d",&a,&b,&c); int k=0; for(int x = 1; x*a < c; x++) { if((c-a*x)%b == 0) k++; } printf("%d\n",k); } return 0; }
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