Nine degree OJ 1083 special multiplication (analog)

Topic 1083: Special multiplication

time limit:1 seconds

Memory limit:32 MB

Special question: No

submitted:2910

Resolution:2027

Title Description:

Write an algorithm that evaluates to 2 inputs that are less than 1000000000.

Special Multiplication Example: 123 * 45 = 1*4 +1*5 +2*4 +2*5 +3*4+3*5

Input:

Two a number less than 1000000000

Output:

Input may have more than one set of data, for each set of data, the output of two numbers in input according to the methods required by the method to calculate the results obtained.

Sample input:

123 45

Sample output:

54

Simple simulation

#include <stdio.h>long long a,b;int x[20];int y[20];void solve () {    int i=0;    int j=0;    int xn=0;    int yn=0;    while (a)    {        x[i++]=a%10;        a/=10;    }    xn=i;    i=0;    while (b)    {        y[i++]=b%10;        b/=10;    }    yn=i;    Long long sum=0;    for (I=0;i<xn;++i) for        (j=0;j<yn;++j)        {            sum+=x[i]*y[j];        }    printf ("%lld\n", sum);} int main (int argc, char *argv[]) {//   freopen ("1083.in", "R", stdin);    while (~scanf ("%lld%lld", &a,&b))    {        solve ();    }    return 0;} /**************************************************************    problem:1083    user:kirchhoff    Language:c    result:accepted    time:0 ms    memory:912 kb*********************************************** *****************/

Nine degree OJ 1083 special multiplication (analog)