玲珑OJ 1083：XJT Love Digits（离线处理+哈希）

Posted 2020-08-27 Shadowdsp

http://www.ifrog.cc/acm/problem/1083

题意：比较好懂。注意答案的x不包含ax本身，所以才输出-1。

思路：离线处理。根据x排序，然后每次更新Hash[]数组就好了。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <string>
 7 #include <iostream>
 8 #include <stack>
 9 #include <map>
10 #include <queue>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 #define N 100010
15 #define INF 0x3f3f3f3f
16 struct P {
17     int x, id;
18 } ask[N];
19 int Hash[N], p[N], ans[N];
20 
21 bool cmpx(const P &a, const P &b) { if(a.x == b.x) return a.id < b.id; return a.x < b.x; }
22 
23 int get(int num) {
24     int ans = 0;
25     while(num) {
26         ans += num % 10;
27         num /= 10;
28     }
29     return ans;
30 }
31 
32 int main()
33 {
34     int t, cas = 1;
35     scanf("%d", &t);
36     while(t--) {
37         int n, q;
38         scanf("%d%d", &n, &q);
39         for(int i = 1; i <= n; i++)
40             scanf("%d", &p[i]);
41         for(int i = 1; i <= q; i++) {
42             scanf("%d", &ask[i].x);
43             ask[i].id = i;
44         }
45         sort(ask + 1, ask + q + 1, cmpx);
46         printf("Case #%d:\n", cas++);
47         memset(Hash, -1, sizeof(Hash));
48         int j = 1, now;
49         for(int i = 1; i <= q; i++) {
50             while(ask[i].x > j && j <= n) {
51                 now = get(p[j]);
52                 Hash[now] = max(Hash[now], p[j]);
53                 j++;
54             }
55             now = get(p[ask[i].x]);
56             if(Hash[now] == -1) ans[ask[i].id] = -1;
57             else ans[ask[i].id] = Hash[now];
58         }
59         for(int i = 1; i <= q; i++) printf("%d\n", ans[i]);
60     }
61     return 0;
62 }