Description Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros). FJ asks that you do this yourself; don't use a special library function for the multiplication. Input * Lines 1..2: Each line contains a single decimal number. Output * Line 1: The exact product of the two input lines Sample Input 11111111111111 1111111111 Sample Output 12345679011110987654321 //一開始沒有考慮邊界情況，導致溢出了 大數乘法模板題 [cpp] #include <stdio.h> #include <string.h> void mult(char a[],char b[],char s[]) { int i,j,k = 0,alen,blen,sum = 0,res[650][650]={0},flag = 0; char result[650]; alen = strlen(a); blen = strlen(b); for(i = 0;i<alen;i++) { for(j = 0;j<blen;j++) res[i][j] = (a[i]-'0')*(b[j]-'0'); } for(i = alen-1;i>=0;i--) { for(j = blen-1;j>=0;j--) { sum = sum+res[i+blen-j-1][j]; } result[k] = sum%10; k++; sum = sum/10; } for(i = blen-2;i>=0;i--) { for(j = 0;j<=i;j++) { sum = sum+res[i-j][j]; } result[k] = sum%10; k++; sum = sum/10; } if(sum) { result[k] = sum; k++; } for(i = 0;i<k;i++) result[i]+='0'; for(i = k-1;i>=0;i--) s[i] = result[k-1-i]; s[k] = '\0'; while(1) { if(strlen(s)!=strlen(a) && s[0] == '0') strcpy(s,s+1); else break; } } int main() { char c[1000],t[1000],sum[100000]; int m; while(~scanf("%s%s",c,t)) { mult(c,t,sum); printf("%s\n",sum); } return 0; }
