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UVa10025 The ? 1 ? 2 ? ... ? n = k problem

2024-09-14 12:06:11   205人阅读

UVa10025

? 1 ? 2 ? ... ? n = k problem

The problem

Given the following formula, one can set operators ‘+‘ or ‘-‘ instead of each ‘?‘, in order to obtain a given k
? 1 ? 2 ? ... ? n = k

For example: to obtain k = 12 , the expression to be used will be:
- 1 + 2 + 3 + 4 + 5 + 6 - 7 = 12 
with n = 7

 

The Input

The first line is the number of test cases, followed by a blank line.

Each test case of the input contains integer k (0<=|k|<=1000000000).

Each test case will be separated by a single line.

The Output

For each test case, your program should print the minimal possible n (1<=n) to obtain k with the above formula.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

 

2

12

-3646397

 

Sample Output

 

7

2701


数学思维：
看似复杂，情况很多种，乍一看摸不着头脑的一道题，让人心生畏惧，Don‘t panic
数学思维，找规律求解
对于0=1+2-3　　3
　　1=-1+2　　 2
　　2=1-2+3　　3
依次查看，每个数的结果好像没什么规律
再看，会发现，若1变为-1，实际上将原数-2
　　　　　　　2变为-2，实际上将原数-4　　　　均为偶数，我们进步了一大步，我感觉离成功不远了，代码铁定不复杂，只需要我们想清楚
　　我们持续+，直到数s>k,此时若(s-k)%2==0,说明，s可将组成s的数中的一部分变为负数来的到k
　　问题转化为求满足s>k&&(s-k)%2==0的数s时，所经历的最大数

注意输出格式！减少不必要的麻烦。

PS：像这种基本的找规律题,数学思维，看得出就看，看不出就试
Version2.0
#include "cstdio"
#include "cstring"
#include "cstdlib"
#include "iostream"
#include "vector"
#include "queue"
using namespace std;
int main()
{
    int t,k,s,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&k);
        if(k==0)
            printf("3\n");
        else
        {
            if(k<0)k=-k;
            s=n=0;
            while(s<k)s+=++n;
            while((s-k)&1)s+=++n;
            printf("%d\n",n);
        }
        if(t)
            printf("\n");
    }
}

 



Version 1.0
#include "cstdio"
#include "cstring"
#include "cstdlib"
#include "iostream"
#include "vector"
#include "queue"
using namespace std;
#define LL long long
int main()
{
    LL k;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&k);
        int sum=0;
        int flag=0;
        if(k==0)
        {
            printf("3\n");
        }
        else
        {
            if(k<0)k=-k;
            int t=1;
            while(1)
            {
                if(sum==k)///全为+ 15
                    break;
                else if(sum>k)
                {
                    if((sum-k)%2==0)///-在前面 13
                        break;
                    if(sum-t==k)///-在最后
                    {
                        flag=1;
                        break;
                    }
                    else if(sum-t>k)
                    {
                        if((sum-t-k)%2==0)///-前后均有  12
                        {
                            flag=1;
                            break;
                        }
                    }
                }
                sum+=t;
                t++;
            }
            if(flag)t++;
            printf("%d\n",--t);
        }
        if(T)printf("\n");
    }
    return 0;
}

 

UVa10025 The ? 1 ? 2 ? ... ? n = k problem


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