[Project Euler] 来做欧拉项目练习题吧: 题目020

[Project Euler] 来做欧拉项目练习题吧: 题目020

周银辉　

题目描述:

n! means n x (n

1) x ... x 3 x 2 x 1

For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

题目分析:

求100!所得结果的各位数字之和

做法可以有两种：

一是做大数乗法，对于Java内置了BigInteger，或者Scheme可以实现无限精度的语言而已，问题就很简单。

否则自己模拟大数乗法，正如下面的multiply。

#include <stdio.h>#include <math.h>#include <stdlib.h>#include <string.h>#define BF_SZ 1000 //buffer size//将整数转换成字符串，并传出长度, 请手动释放返回值//stdlib中的itoa也能完成类似的功能，//但stdlib传入的buffer长度未知，一般会设置得比较大，而造成浪费char* int_to_str(int n, int* length){ *length = floor(log10(abs(n!=0?n:1))) + 1; char *str = (char*)calloc(sizeof(char)*(*length+1)); int i = *length-1; while(n!=0) { str[i] = n%10+48; n = n/10; i--; } return str;}//大数乗法，结果放在r中void multiply(const char *a,const char *b, char *r){ int i, j, *t, length_a, length_b, length_t; // t for temp length_a = strlen(a); length_b = strlen(b); length_t = length_a + length_b; t=(int*)malloc(sizeof(int)*length_t); for(i=0; i<length_t; i++) { t[i]=0; } for (i=0; i<length_a; i++) { for (j=0;j<length_b;j++) { t[i+j+1] += (a[i]-48)*(b[j]-48); } } // 这里实现进位操作 for (i=length_t-1; i>=0; i--) { if(t[i]>=10) { t[i-1] += t[i]/10; t[i] %= 10; } } i=0; while(t[i]==0) { i++; // 跳过数前面的0 } for (j=0; i<length_t; i++,j++) { r[j] = t[i]+48; } r[j]='\0'; free(t); }int test(int n, char* buffer, char* temp){ int i, j, k, a, b, c, p; //p for product int len; for(i=2; i<=n; i++) { char *str = int_to_str(i, &len); //printf("%s\t", str); //做buffer和str的大数乘法 multiply(buffer, str, buffer); free(str); } return 0;}int main(){ char* buffer = (char*)malloc(BF_SZ*sizeof(char)); char* temp = (char*)malloc(BF_SZ* sizeof(char)); memset(buffer, '0', BF_SZ); buffer[BF_SZ-1] = '1'; memset(temp, '0', BF_SZ); test(100, buffer, temp); printf("buffer is %s\n", buffer); int i, len=strlen(buffer), count=0; for (i=0; i<len; i++) { count += buffer[i]-48; } printf("the count is %d\n", count); free(buffer); free(temp); return 0;}

二是做大数阶乘：#include <stdio.h>#include <stdlib.h>int buffer[1000];void multiply(int buffer[],int n,int *length){ int c=0, i, len=*length; for(i=0; i<len; i++) { buffer[i] = buffer[i]*n + c; c = buffer[i]/10; buffer[i]%=10; } while(c!=0) { buffer[len++] = c%10; c /= 10; } *length=len;}main(){ buffer[0]=1; int i, n=100, len=1; for(i=2; i<=n; i++) { multiply(buffer, i, &len); } for(i=len-1; i>=0; i--) { printf("%d",buffer[i]); } printf("\n");}

这种模拟大数运算的题目要求细心和耐性,其它便没什么了