HDU-1002.大数相加(字符串模拟)

  本题大意:给出两个1000位以内的大数a 和b,让你计算a + b的值。

  本题思路:字符串模拟就能过,会Java的大佬应该不会点进来......

  参考代码:

 1 #include <cstdio>
 2 #include <cstring>
 3 using namespace std;
 4 
 5 const int maxn = 1000 + 5;
 6 int  t, Case = 0, now;
 7 int ans[maxn];
 8 char s1[maxn], s2[maxn];
 9 
10 int main () {
11     scanf("%d", &t);
12     while(t --) {
13         if(Case > 0) printf("\n");
14         now = 0;
15         memset(ans, 0, sizeof(ans));
16         scanf("%s %s", s1, s2);
17         int len1 = strlen(s1), len2 = strlen(s2), i = len1 - 1, j = len2 - 1;
18         while(i >= 0 && j >= 0) {
19             ans[now + 1] =  (ans[now] + s1[i] + s2[j] - 2 * '0') / 10;
20             ans[now] = (ans[now ++] + s1[i --] + s2[j --] - 2 * '0') % 10;
21         }
22         while(i >= 0) {
23             ans[now + 1] = (ans[now] + s1[i] - '0') / 10;
24             ans[now] = (ans[now ++] + s1[i --] - '0') % 10;
25         }
26         while(j >= 0) {
27             ans[now + 1] = (ans[now] + s2[j] - '0') / 10;
28             ans[now] = (ans[now ++] + s2[j --] - '0') % 10;
29         }
30         printf("Case %d:\n%s + %s = ", ++ Case, s1, s2);
31         if(ans[now] > 0)    now ++;
32         for(i = now - 1; i >= 0; i --)
33             printf("%d", ans[i]);
34         printf("\n");
35     }
36     return 0;
37 }
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posted @ 2019-03-09 16:09  Cruel_King  阅读(177)  评论(0编辑  收藏  举报