Wiki oi3117 high-precision multiplication of exercises

Title Description Description

Give two positive integers a and B to calculate the value of the a*b. Ensure that the number of digits A and B does not exceed 500 digits.

Enter a description Input Description

Read in two positive integers separated by a space

Output description Output Description

The value of the output a*b

Sample input Sample Input

3 12

Sample output Sample Output

36

Data range and Tips Data Size & Hint

The number of digits of two positive integers is less than 500 bits

Code:

1#include <stdio.h>2#include <string.h>3#include <math.h>4#include <iostream>5#include <algorithm>6 using namespacestd;7 Chara[ -],b[ -];8 intc[502],d[ the];//C Save Intermediate data, D keep accumulating C answer9 intMain ()Ten { OneMemset (c,0,sizeof(c)); Amemset (D,0,sizeof(d)); -Cin>>a>>b; -     intAlen=strlen (a)-1; the     intBlen=strlen (b)-1; -      for(intI=alen; i>=0; i--) -     { -         intcarry=0, carry1=0, k=0, m=0;//K as subscript for C, M as subscript for D, carry for C (multiply), and carry1 for D (cumulative) +          for(intJ=blen; j>=0; j--) -         { +             intCc= (a[i]-'0') * (b[j]-'0')+carry; Ac[k++]=cc%Ten; atcarry=cc/Ten; -         } -         if(carry!=0) C[k++]=carry;//note that after the last one, there may be rounding . - //for (int n=k-1; n>=0; n--) printf ("%d", C[n]);//View Intermediate data for C - //printf ("\ n"); -m=alen-i; in          for(intL=0; l<k; l++) -         { to             intCc= (D[m]+c[l]) +Carry1; +d[m++]=cc%Ten; -carry1=cc/Ten; the         } *         if(carry1!=0) D[m++]=carry1;//note that after the last one, there may be rounding . $ //for (int n=m-1; n>=0; n--) printf ("%d", D[n]);Panax Notoginseng //printf ("\ n"); -     } the          for(intn=m-1; n>=0; n--) printf ("%d", D[n]); +printf"\ n"); A     return 0; the}

Wiki oi3117 high-precision multiplication of exercises