Power Calculus
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 1760 | Accepted: 947 |
Description
Starting with x and repeatedly multiplying by x, we can computex31 with thirty multiplications:
x2 = x × x, x3 =
x2 × x, x4 = x3 ×x, …,
x31 = x30 × x.
The operation of squaring can be appreciably shorten the sequence of multiplications. The following is a way to computex31 with eight multiplications:
x2 = x × x, x3 =
x2 × x, x6 = x3 ×x3,
x7 = x6 × x,x14 =
x7 × x7, x15 =x14 ×
x, x30 = x15 ×x15,
x31 = x30 × x.
This is not the shortest sequence of multiplications to compute x31. There are many ways with only seven multiplications. The following is one of them:
x2 = x × x, x4 = x2 ×x2,
x8 = x4 × x4,x8 =
x4 × x4, x10 =x8 ×
x2, x20 = x10 ×x10,
x30 = x20 × x10,
x31 = x30 × x.
If division is also available, we can find a even shorter sequence of operations. It is possible to computex31 with six operations (five multiplications and one division):
x2 = x × x, x4 =
x2 × x2, x8 = x4 ×x4,
x16 = x8 × x8,x32 =
x16 × x16, x31 =
x32 ÷ x.
This is one of the most efficient ways to compute x31 if a division is as fast as a multiplication.
Your mission is to write a program to find the least number of operations to computexn by multiplication and division starting with
x for the given positive integern. Products and quotients appearing in the sequence should be
x to a positive integer’s power. In others words,x?3, for example, should never appear.
Input
The input is a sequence of one or more lines each containing a single integer
n. n is positive and less than or equal to 1000. The end of the input is indicated by a zero.
Output
Your program should print the least total number of multiplications and divisions required to computexn starting with
x for the integer n. The numbers should be written each in a separate line without any superfluous characters such as leading or trailing spaces.
Sample Input
1 31 70 91 473 512 811 953 0
Sample Output
0 6 8 9 11 9 13 12 代码:
#include<iostream>
#include<cstdio>
using namespace std;
int T;
const int mod = 1e9+7;
long long int pow(int num,int coun)
{
long long int p=num,q=coun,ret=1;
while(q)
{
if(q & 1) {
ret=ret*p%mod;
}
q>>=1;
p=p*p%mod;
}
return ret;
}
int main(){
scanf("%d",&T);
while(T--){
int n,k;
scanf("%d%d",&n,&k);
if(n==k){printf("1\n");continue;}
if(n<k){printf(("0\n"));continue;}
if(n==k+1){printf("2\n");continue;}
printf("%I64d\n",pow(2,(n-k-2))*(n-k+3)%mod);
}
}