HDU 4952 Number Transformation 打表规律

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Number Transformation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 495 Accepted Submission(s): 248

Problem Description

Teacher Mai has an integer x.

He does the following operations k times. In the i-th operation, x becomes the least integer no less than x, which is the multiple of i.

He wants to know what is the number x now.

Input

There are multiple test cases, terminated by a line "0 0".

For each test case, the only one line contains two integers x,k(1<=x<=10^10, 1<=k<=10^10).

Output

For each test case, output one line "Case #k: x", where k is the case number counting from 1.

Sample Input

2520 102520 200 0

Sample Output

Case #1: 2520Case #2: 2600

Source

2014 Multi-University Training Contest 8

给你一个数x和k次操作，要求没次操作使得x*n能整除i，并且x*n是最接近x的数，求最后的数。

(i+1)*x'>=i*x => x'>=i*x/(i+1) => x'>=x-x/(i+1) 当x <=i+1的时候，x'就不变了，所以暴力求解就可以了。

//93MS256K#include<stdio.h>#include<string.h>#define ll __int64ll a,b;int main(){    int cas=1;    while(scanf("%I64d%I64d",&a,&b),a|b)    {        ll i,c;        int flag=0;        for(i=1;i<=b;i++)        {            if(a%i)            {                c=a/i+1;a=c*i;                //printf("a=%I64d,c=%I64d,i=%I64d\n",a,c,i);                if(c<=i){flag=1;break;}            }        }        if(flag)printf("Case #%d: %I64d\n",cas++,c*b);        else printf("Case #%d: %I64d\n",cas++,a);    }}

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