Codeforce - 267 - A. Subtractions

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本文介绍了一种算法，该算法通过不断将较大数减去较小数的方式，直至其中一个数变为0，以此来计算操作次数。文章提供了一个AC代码示例，展示了如何使用C++实现这一过程。

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A. Subtractions
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You’ve got two numbers. As long as they are both larger than zero, they go through the same operation: subtract the lesser number from the larger one. If they equal substract one number from the another. For example, one operation transforms pair (4,17) to pair (4,13), it transforms (5,5) to (0,5).

You’ve got some number of pairs (ai, bi). How many operations will be performed for each of them?

Input
The first line contains the number of pairs n (1 ≤ n ≤ 1000). Then follow n lines, each line contains a pair of positive integers ai, bi (1 ≤ ai, bi ≤ 109).

Output
Print the sought number of operations for each pair on a single line.

Examples
input
2
4 17
7 987654321
output
8
141093479

题意：大数减小数，直到其中一个数为 0 为止。求步数。

AC代码：

#include <iostream>
using namespace std;

int main()
{
    int n;
    cin>>n;
    while (n--)
    {
        int a,b;
        cin>>a>>b;
        int ans=0;
        int mn=min(a,b),mx=max(a,b);
        int ansmn=mn,ansmx=mx;
        do
        {
            ans+=(mx/mn);
            mx%=mn;
            ansmn=min(mx,mn);ansmx=max(mx,mn);
            mn=ansmn;mx=ansmx;
        }while (mx && mn);
        cout<<ans<<endl;
    }
}