Steps POJ - 2590
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
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
4
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);
}
}
}