UC知道谁帮我做acm题，高手帮忙啊
来源：百度知道 编辑：UC知道 时间：2024/07/05 00:12:02
Combinations
Description
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
Input
The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M p
Description
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
Input
The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M p
#include <stdio.h>
int main ()
{
int a[2000],d=3,i,j,b,c,x,f,y;
while(1)
{
scanf("%d %d",&b,&c);
if(b==0&&c==0)break;
printf("%d things taken %d at a time is ",b,c);
memset(a,0,sizeof(a));
a[0]=1;
for(i=b;i>b-c;i--)
{
for(j=0,f=0;j<d;j++)
{
x=f+a[j]*i;
f=x/10;
a[j]=x%10;
}
d+=2;
}
while(a[j]==0)j--;
y=j;
d=j;
for(i=c;i>1;i--)
{
for(f=0;j>=0;j--)
{
x=f*10+a[j];
a[j]=x/i;
f=x%i;
}
j=d;
}
while(a[y]==0)y--;
for(i=y;i>=0;i--)
printf("%d",a[i]);
printf(" exactly.\n");
}
return 0;
}
AC的.
题目还可以的.不用注释了吧.
我可以帮，1题1元。