CodeForces 792C - Divide by Three [ 分类讨论 ]

删除最少的数位和前缀0，使得剩下的数能被3整除

等价于各数位数字之和能被3整除。

当前数位和可能是 0, 1, 2(mod 3)

　　0： 直接处理

　　1： 删除一个a[i]%3 == 1 或者 两个a[i]%3 == 2

　　2: 同1

对于删完的数列，去掉前置0（只剩前置0就当作0）

若删啥都不满足，则判断原数列中有没有0，不然就输出-1

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 100005;
 4 char s[MAXN];
 5 int a[MAXN], n;
 6 string ans, tmp;
 7 int b[MAXN], m;
 8 void Update()
 9 {
10     tmp.clear();
11     int i;
12     for (i = 1; i <= m && !b[i]; i++);
13     if (i < m+1)
14     {
15         for (i; i <= m; i++)
16             tmp += b[i] + ‘0‘;
17     }
18     else if (m)
19         tmp += ‘0‘;
20     if (ans.length() < tmp.length())
21         ans = tmp;
22 }
23 int main()
24 {
25     scanf("%s", s);
26     n = strlen(s);
27     for (int i = 1; i <= n; i++)
28         a[i] = s[i-1] - ‘0‘;
29     int sum = 0;
30     for (int i = 1; i <= n; i++)
31         sum = (sum+a[i]%3) % 3;
32     if (!sum)
33     {
34         m = 0;
35         for (int i = 1; i <= n; i++)
36             b[++m] = a[i];
37         Update();
38     }
39     else
40     {
41         int p1 = 0, p2 = 0;
42         for (int i = n; i > 0 && !p1; i--)
43             if (a[i]%3 == sum) p1 = i;
44         if (p1)
45         {
46             m = 0;
47             for (int i = 1; i <= n; i++)
48             {
49                 if (i == p1) continue;
50                 b[++m] = a[i];
51             }
52             Update();
53         }
54         p1 = p2 = 0;
55         for (int i = n; i > 0 && (!p1 || !p2) ; i--)
56             if (a[i]%3 +sum == 3) p1 ? p2 = i : p1 = i;
57         if (p1 && p2)
58         {
59             m = 0;
60             for (int i = 1; i <= n; i++)
61             {
62                 if (i == p1 || i == p2) continue;
63                 b[++m] = a[i];
64             }
65             Update();
66         }
67     }
68     if (!ans.length())
69     {
70         bool flag = 0;
71         for (int i = 1; i <= n; i++)
72             if (!a[i]) flag = 1;
73         puts(flag? "0": "-1");
74     }
75     else cout << ans << endl;
76 }