2012-12-24 17:11:44Morris
[UVA] 11403 - Binary Multiplication
Problem: I
Binary Multiplication
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All of us know simple multiplication rule. Can we make a program to multiply two binary numbers? Well we can try.
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Input specification:
Each case will define two binary strings (x & y). You can assume that length of each string will not exceed 30. The program will terminate with a 0 0.
Out put specification:
Out put must be formatted like the following examples. You have to show the step by step procedure for the multiplication. Each result will be separated by an empty line.
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Sample Input: |
Output for Sample Input: |
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11 11 111 10 10 111 0 0 |
11 11 -- 11 11 ---- 1001
111 10 --- 000 111 ---- 1110
10 111 --- 10 10 10 ---- 1110 |
這題完全不知道要搞什麼鬼，一堆錯誤題意。
#include <stdio.h>
#include <string.h>
int main() {
char a[105], b[105];
int first = 0;
while(scanf("%s %s", a, b) == 2) {
if(!strcmp("0", a) && !strcmp("0", b))
break;
int la = strlen(a), lb = strlen(b);
int ans[105][105] = {};
int i, j, ia, ja;
for(i = lb-1, ia = 0; i >= 0; i--, ia++) {
for(j = la-1, ja = 0; j >= 0; j--, ja++) {
ans[ia][ia+ja] += (b[i]-'0')*(a[j]-'0');
ans[lb][ia+ja] += ans[ia][ia+ja];
}
for(j = 0; j < 100; j++) {
ans[i][j+1] += ans[i][j]/2;
ans[i][j] %= 2;
}
for(j = 0; j < 100; j++) {
ans[lb][j+1] += ans[lb][j]/2;
ans[lb][j] %= 2;
}
}
int mx = la > lb ? la : lb;
int anslen;
for(i = 99; i >= 0; i--)
if(ans[lb][i]) break;
if(i < 0) i = 0;
anslen = i+1;
if(i+1 > mx) mx = i+1;
for(i = 0; i < mx-la; i++)
putchar(' ');
puts(a);
for(i = 0; i < mx-lb; i++)
putchar(' ');
puts(b);
int mx2 = la > lb ? la : lb;
for(i = 0; i < mx-mx2; i++)
putchar(' ');
for(i = 0; i < mx2; i++)
putchar('-');
puts("");
for(i = 0; i < lb; i++, puts("")) {
for(j = 0; j < mx-i-la; j++)
putchar(' ');
for(j = i+la-1; j >= i; j--)
putchar(ans[i][j]+'0');
}
for(i = 0; i < mx; i++)
putchar('-');
puts("");
for(i = 0; i < mx-anslen; i++)
putchar(' ');
for(i = anslen-1; i >= 0; i--)
putchar(ans[lb][i]+'0');
puts("");
puts("");
}
return 0;
}
/*
11 11
111 10
10 111
0 0
*/