Russian乗算のプログラム ref: http://qiita.com/items/8317785118b1f8091818

russian.c

#include <stdio.h>

int Russian (int x, int y){
    int m;

    if (x % 2 != 0) {
        m = y;
    } else {
        m = 0;
    }

    while (x != 1) {
        x = x / 2;
        y = 2 * y;
        if (x % 2 != 0) {
            m = m + y;
        }
    }

    return m;
}

int main(int argc, const char * argv[])
{
    int a = 36, b = 17, result = Russian(a,b);
    printf("%d × %d = %d\n",a, b, result);
    return 0;
}