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

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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;
}

 

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