Steps POJ - 2590

最新推荐文章于 2021-12-03 14:53:15 发布

转载最新推荐文章于 2021-12-03 14:53:15 发布 · 106 阅读

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原文链接：http://www.cnblogs.com/da-mei/p/9053229.html

本文介绍了一种计算从整数点x到y所需的最小行走步数的算法，采用贪心策略，将问题转化为寻找特定序列并进行优化。

Steps
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8219 Accepted: 3747
Description

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

Input consists of a line containing n, the number of test cases.
Output

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.
Sample Input

3
45 48
45 49
45 50
Sample Output

3
3

贪心做法，先把 x写成1+2+3+.....+n+n-1+n-2+n-3+....+2+1的形式，再把剩下的放进去。

上面的式子等于 n^2,所以先判断x是否为平方数，如果是的话直接输出 2*k-1，如果不是在加上剩下。

#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define maxn 1100000
#define maxm 1000000000005
#define mod 1000000007
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
     ll x,y;scanf("%lld%lld",&x,&y);
     if(x==y){
        printf("0\n");continue;
     }
     ll k=sqrt((y-x)*1.0);
     if(k*k==y-x)printf("%lld\n",2*k-1);
     else {
        ll ans=(y-x)-k*k;
        ll ans1=2*k-1;
        while(ans!=0){
           ans1+=(ans/k);
           ans%=k;
           k--;
        }
        printf("%lld\n",ans1);
     }
    }
}

转载于:https://www.cnblogs.com/da-mei/p/9053229.html