洛谷P1303P1303 A*B Problem - 洛谷 | 電腦科學教育新生態 (luogu.com.cn)

方法：陣列模擬

 1 //#include <bits/stdc++.h>
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <cstring>
 6 #include <string>
 7 #include <cmath>
 8 #include <queue>
 9 #include <stack>
10
 
 #include <vector>
11 #define INF 0x3f3f3f3f
12 const int maxn = 250;
13 typedef long long ll;
14 
15 using namespace std;
16 
17 char s1[2005], s2[2005];
18 int a[5005], b[5005], c[5006];
19 int la, lb, lc;
20 int main() {
21     scanf("%s", s1);
22     scanf("%s", s2);
23     la = strlen(s1);
24     lb = strlen(s2);
 
25     lc = la + lb; //注意 
26     for(int i = 0; i < la; i++){
27         a[la - i] = s1[i] - '0';
28             
29     } 
30     for(int i = 0; i < lb; i++){
31         b[lb - i] = s2[i] - '0';
32     }
33     for (int i = 1;  i <= la; i++){
34         for (int j = 1; j <= lb; j++){
35             c[i + j - 1
 
] += a[i]*b[j];
36             c[i + j] += c[i + j - 1] / 10;
37             c[i + j - 1] %= 10;
38         }
39     }
40     
41 //        if(c[lc] == 0 && lc > 0) lc--; //當出現10000 * 0 時會出現錯誤 
42     while(c[lc] == 0 && lc > 1){
43         lc--;
44     }
45     for (int i = lc; i > 0; i--){
46         printf("%d", c[i]);
47     }
48 
49     return 0;
50 }