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  • BestCoder Round #75 King's Cake 模拟&&优化 || gcd

    King's Cake

    Accepts: 967
    Submissions: 1572
    Time Limit: 2000/1000 MS (Java/Others)
    Memory Limit: 65536/65536 K (Java/Others)
    Problem Description

    It is the king's birthday before the military parade . The ministers prepared a rectangle cake of size n×m(1≤n,m≤10000) . The king plans to cut the cake himself. But he has a strange habit of cutting cakes. Each time, he will cut the rectangle cake into two pieces, one of which should be a square cake.. Since he loves squares , he will cut the biggest square cake. He will continue to do that until all the pieces are square. Now can you tell him how many pieces he can get when he finishes.

    Input

    The first line contains a number T(T≤1000), the number of the test cases.

    For each test case, the first line and the only line contains two positive numbers n,m(1≤n,m≤10000).

    Output

    For each test case, print a single number as the answer.

    Sample Input
    2
    2 3
    2 5
    Sample Output
    3
    4
    
    hint:
    For the first testcase you can divide the into one cake of 2×2 , 2 cakes of 1×1

    source

    The question is from BC and hduoj 5640.


    My Solution

    //A really easy problem, I get a Runtime Error(STACK_OVERFLOW) first, then Time Limit Exceeded
    //next Runtime Error (INTEGER_DIVIDE_BY_ZERO), and a WA   , Accepted......
    //I am really sorry for that.

    1、用循环模拟

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    int tot;
    
    int main()
    {
        int T, l, s, t;
        scanf("%d", &T);
        while(T--){
            scanf("%d%d", &l, &s);
            if(l < s) swap(l, s);
            tot = 0;
            while(true){
                if(s == 1) {tot += l; break;}
                if( s == 0) break;               //!
                t = l/s;
                l -= t*s;
                if(l < s) swap(l, s);
                tot += t;
            }
    
            printf("%d
    ", tot);
        }
        return 0;
    }


    2、像是求最大公约数。所以每次 gcd 的时候累加答案就可以,复杂度O(logmax(n,m)T)


    Thank you!


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  • 原文地址:https://www.cnblogs.com/zsychanpin/p/7116418.html
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