九度1076 N的阶乘

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n数值较小，可以直接与每位上的数相乘。

#include <stdio.h>#include <stdlib.h>#include <string.h>void reverse(char * s){    char temp;    int i,j;    for(i=0,j=strlen(s)-1;i<j;i++,j--)    {        temp=s[i];        s[i]=s[j];        s[j]=temp;    }}int main(){    int n;    char res[3000];    int i,j;    int add;    int len;    while(~scanf("%d",&n))    {        add=0;        len=1;        memset(res,'0',3000);        res[0]='1';        for(i=2;i<=n;i++)        {            for(j=0;j<len;j++)            {                add += (res[j]-'0')*i;                res[j] = add%10 + '0';                add /= 10;            }            for(;add!=0;add/=10)              res[len++]=add%10+'0';        }        res[len]='\0';        reverse(res);        printf("%s\n",res);    }    return 0;}

用大数乘法实现

#include <stdio.h>#include <stdlib.h>#include <string.h>void reverse(char * s){    char temp;    int i,j;    for(i=0,j=strlen(s)-1;i<j;i++,j--)    {        temp=s[i];        s[i]=s[j];        s[j]=temp;    }}void multi(char *lef,char * rig,char * res){    int ll,rl;    int i,j;    int add,mul;    ll = strlen(lef);    rl = strlen(rig);    int len;    int temp1,temp2;    for(i = 0 ; i < rl ; i ++)    {        add=mul=0;        for(j=0;j<ll;j++)        {            temp1 = (lef[j]-'0')*(rig[i]-'0') + mul;            mul = temp1/10;            temp1 %= 10;            if(res[i+j]==0)            {                temp2= temp1+add;            }            else            {                temp2 = res[i+j] - '0' + temp1 + add;            }            add = temp2 / 10;            res[i+j] = temp2%10 + 48;        }        len = i + ll;        for( ; add != 0 || mul!=0 ; add /= 10,mul /= 10)        {            res[len++] = '0' + mul%10 + add%10;        }    }}int main(){    int n;    char res[3000];    char right[5];    char left[3000];    int i,j;    int add;    int len;    while(~scanf("%d",&n))    {        add=0;        memset(left,0,3000);        memset(res,0,3000);        left[0]='1';        for(i=2;i<=n;i++)        {            /*for(j=0;j<len;j++)            {                add += (res[j]-'0')*i;                res[j] = add%10 + '0';                add /= 10;            }            for(;add!=0;add/=10)              res[len++]=add%10+'0';*/            memset(res,0,3000);            sprintf(right,"%d",i);            reverse(right);            multi(left,right,res);            strcpy(left,res);        }        reverse(res);        printf("%s\n",res);    }    return 0;}

简单的大数乘法实现

void multiply(const char *a,const char *b){     int i,j,ca,cb,*s;     ca=strlen(a);     cb=strlen(b);     s=(int *)malloc(sizeof(int)*(ca+cb));   //分配存储空间     for (i=0;i<ca+cb;i++) s[i]=0;      // 每个元素赋初值0          for (i=0;i<ca;i++)         for (j=0;j<cb;j++)             s[i+j+1]+=(a[i]-'0')*(b[j]-'0');                  for (i=ca+cb-1;i>=0;i--)        // 这里实现进位操作         if (s[i]>=10)         {             s[i-1]+=s[i]/10;              s[i]%=10;         }          char *c=(char *)malloc((ca+cb)*sizeof(char));  //分配字符数组空间，因为它比int数组省！     i=0;while(s[i]==0) i++;   // 跳过头部0元素     for (j=0;i<ca+cb;i++,j++) c[j]=s[i]+'0';     c[j]='\0';     for (i=0;i<ca+cb;i++) cout<<c[i];     cout<<endl;      free(s);}

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