lightoj 1282 - Leading and Trailing (数学--log使用)
2016-08-23 20:56
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1282 - Leading and Trailing
You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of nk.
Each case starts with a line containing two integers: n (2 ≤ n < 231) and k (1 ≤ k ≤ 107).
题意:问N^K的前三位和后三位(要补0处理)
思路:后三位:快速幂取模
前三位技巧:
N^K = X *10 ^(len-1) //len是整数的长度
两边取对数:K*log10(N)=(len-1)+log( X )
//求出X
总结:tmp=k*log10(n)
tmp=tmp-[tmp]
tmp=tmp*100
代码:
#include<cstdio>
#include<cmath>
#define LL long long
LL powmod(LL x,LL n)//快速幂
{
LL res=1;
while(n>0)
{
if(n&1)
res=res*x%1000;
x=x*x%1000;
n>>=1;
}
return res;
}
int main()
{
int t,kcase=1,m=100;
LL n,k;
LL s1,s2;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&n,&k);
s1=powmod(n,k);
double tmp=k*log10(n*1.0);
tmp=tmp-(long long)tmp;
tmp=pow(10.0,tmp);
printf("Case %d: %lld %03lld\n",kcase++,(LL)(tmp*100),s1); //补0处理
}
return 0;
}
PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
Input
Input starts with an integer T (≤ 1000), denoting the number of test cases.Each case starts with a line containing two integers: n (2 ≤ n < 231) and k (1 ≤ k ≤ 107).
Output
For each case, print the case number and the three leading digits (most significant) and three trailing digits (least significant). You can assume that the input is given such that nk contains at least six digits.Sample Input | Output for Sample Input |
5 123456 1 123456 2 2 31 2 32 29 8751919 | Case 1: 123 456 Case 2: 152 936 Case 3: 214 648 Case 4: 429 296 Case 5: 665 669 |
思路:后三位:快速幂取模
前三位技巧:
N^K = X *10 ^(len-1) //len是整数的长度
两边取对数:K*log10(N)=(len-1)+log( X )
//求出X
总结:tmp=k*log10(n)
tmp=tmp-[tmp]
tmp=tmp*100
代码:
#include<cstdio>
#include<cmath>
#define LL long long
LL powmod(LL x,LL n)//快速幂
{
LL res=1;
while(n>0)
{
if(n&1)
res=res*x%1000;
x=x*x%1000;
n>>=1;
}
return res;
}
int main()
{
int t,kcase=1,m=100;
LL n,k;
LL s1,s2;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&n,&k);
s1=powmod(n,k);
double tmp=k*log10(n*1.0);
tmp=tmp-(long long)tmp;
tmp=pow(10.0,tmp);
printf("Case %d: %lld %03lld\n",kcase++,(LL)(tmp*100),s1); //补0处理
}
return 0;
}
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