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  • Hdoj 2058

    原题链接

    描述

    Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.

    输入

    Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

    输出

    For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

    样例输入

    20 10
    50 30
    0 0

    样例输出

    [1,4]
    [10,10]

    [4,8]
    [6,9]
    [9,11]
    [30,30]

    思路

    等差数列求和的变形。
    (S=frac{(k+k+n-1)*n}{2})
    (2k=frac{2s}{n}-n+1, n≤sqrt{S})
    枚举n然后算出k就好

    代码

    #include <bits/stdc++.h>
    #define ll long long
    using namespace std;
    
    int main()
    {
    	ll n, m;
    	while(~scanf("%lld %lld", &n, &m))
    	{
    		if(n + m == 0) break;
    		ll k, t;
    		for(t = sqrt(2 * m); t > 0; t--)
    		{
    			if(2 * m % t) continue;
    			ll r = 2 * m / t - t + 1;
    			if(r % 2) continue;
    			k = r >> 1;
    			printf("[%lld,%lld]
    ", k, k + t - 1);
    		}
    		printf("
    ");
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/HackHarry/p/8371005.html
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