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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.
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).
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 10 2520 20 0 0
Sample Output
Case #1: 2520 Case #2: 2600
Source
2014 Multi-University Training Contest 8
(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。
//93MS 256K #include<stdio.h> #include<string.h> #define ll __int64 ll 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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