Problem E - Steps（問題e-步驟）

題目描述

One steps through integer points of the straight line. The length of a step must be nonnegative and can be by one bigger than, equal to, or by one smaller than the length of the previous step.

What is the minimum number of steps in order to get from x to y? The length of the first and the last step must be 1.

Input consists of a line containing n, the number of test cases. For each test case, a line follows with two integers: 0 <= x <= y < 2^31. For each test case, print a line giving the minimum number of steps to get from x to y.

樣例輸入

3
45 48
45 49
45 50

樣例輸出

3
3
4

題意：

一個步驟通過直線的整數點。一個步驟的長度必須是非負的，並且可以由一個大於、等於或小於前一個步驟的長度。

從X到Y的最小步數是多少？第一步和最後一步的長度必須是1。

輸入由一行組成，其中包含N、測試用例的數目。對於每個測試用例，一行後面有兩個整數：0<=x<=y<2^31.對於每個測試用例，列印一行，給出從X到Y的最小步驟數。

AC碼：

#include<stdio.h>
int main()
{
	int n;
	long long x,y;
	long long di;
	int i,d;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%lld %lld",&x,&y);
		di=y-x;
		for(i=d=0;i+d<di;i++)
		d+=i/2;
		printf("%d\n",i);
	}
	return 0;
}

#include<stdio.h>
 
int main(){
	long long a, b, t, n, sum, k;
	scanf("%lld", &t);
	while (t--){
		scanf("%lld%lld", &a, &b);
		n = b - a;
		sum =  0;
		if (n == 0){
			printf("%lld\n", n);
			continue;}
 
		for (long long i = 0; ; i++){
			sum += 2 * i;
			k = sum - n;
			if (k >= 0 && k < i){
				printf("%lld\n", 2 * i);
				break;
			}
			else if(k >= i){
				printf("%lld\n", 2 * i - 1);
				break;
			}
		}
	}
	return 0;}
/*題目大意：給出兩個數字， 要求從a走到b, 第一步和最後一步只能走1， 每步的值可以是前面一步值-1， 不變和+1.求a到b最少走幾步。
解題思路：1...k k ...1所需步為2 * k,

<1>2 * sum(1~k)  ->2 * k

<2>2 * sum(1~k)  ~2 *  sum(1~k) + 1 + k  -> 2 *k + 1

<3>2 * sum(1~k) + k + 1 ~2 * sum(1~k+1)  -> 2 *(k + 1)
*/

#include<stdio.h>
int main()
{
    int n,x,y;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d",&x,&y);
        int mid=(y-x)/2;
        int i=0,sum=0;
        while(1)
        {
 
            if(sum<=mid)
            {
                i++;
                sum+=i;
            }
            else
           {
               sum=sum-i;
               i=i-1;
                break;
           }
        }
        //printf("i=%d sum=%d\n",i,sum);
        //printf("%d %d\n",x,y);
        if(y-x-sum*2==0)
        i=i*2;
        else if((y-x-sum*2)<=i||(y-x-sum*2)==i+1)
        i=i*2+1;
        else
        i=(i+1)*2;
        printf("%d\n",i);
    }
    return 0;
}
/*首次、末次都是1，中間數只能比前一個數大一，或小一、或相等*/