Judge Info
- Memory Limit: 32768KB
- Case Time Limit: 10000MS
- Time Limit: 10000MS
- Judger: Normal
Description
Please calculate the answer of A*B, both A and B are integer.
Input
The first line of input contains 
, the number of test cases. There is only line for each test case. It contains two integers 
.
Output
For each test case, output A*B in one line.
Sample Input
3
1 2
1 1
-1 -1
Sample Output
2
1
1
Ps : 算法课的小练习，通过字符串处理。注意0就OK了。
#include <stdio.h>
#include <string.h> char A[];
char B[];
int C[++]; char *revstr(char *str, int len){
char *start = str;
char *end = str + len - ;
char ch;
if(str != ){
while(start < end){
ch = *start;
*start ++ = *end;
*end-- = ch;
}
}
return str;
} int main(int argc, char const *argv[])
{
int t, i, j, lenA, lenB, len, Amark, Bmark, mark;
scanf("%d", &t);
while(t--){ mark=;
Amark = ;
Bmark = ; memset(A, '', sizeof(A));
memset(B, '', sizeof(B));
memset(C, , sizeof(C));
scanf("%s", A);
scanf("%s", B);
lenA = strlen(A);
lenB = strlen(B); if(A[] == '-'){
for(i=;i<lenA;++i)
A[i-] = A[i];
A[i-]= '\0';
A[i] = '';
Amark = ;
lenA--;
} if(B[] == '-'){
for(i=;i<lenB;++i)
B[i-] = B[i];
B[i-] = '\0';
B[i] = '';
Bmark = ;
lenB--;
}
revstr(A,lenA);
revstr(B,lenB);
A[lenA] = '\0';
B[lenB] = '\0'; for(i=;i<lenB;++i){
for(j=;j<lenA;++j){
C[j+i]+=(B[i]-'')*(A[j]-'');
}
}
len = lenA + lenB;
for(i=;i<len;++i){
if(C[i] > ){
C[i+]+=C[i]/;
C[i] %= ;
}
}
while(!C[len-]){
len--;
}
//printf("len = %d\n", len);
if(len == ){ printf("0\n"); continue;}
if(len > - && Amark != Bmark)
putchar('-');
for(i=len-;i>=;i--)
printf("%d", C[i]); printf("\n");
} return ;
}