Codeforces Round #213 (Div. 2) 解题报告

Problem A Good Number

找出数组中包含所有小于等于k的数字的数。水题

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <cstdlib>
 6 #include <algorithm>
 7 
 8 using namespace std;
 9 
10 int n, k, vis[11];
11 
12 bool judge(long long num)
13 {
14     memset(vis, 0, sizeof vis);
15     while(num){
16         vis[num%10] = 1;
17         num/=10;
18     }
19     for(int i=0; i<=k; i++){
20         if(vis[i] == 0)return false;
21     }
22     return true;
23 }
24 
25 int main()
26 {
27 //    freopen("in.txt", "r", stdin);
28 
29     long long num;
30     while(scanf("%d%d", &n, &k)!=EOF){
31         int ans = 0;
32         for(int i=0; i<n; i++){
33             scanf("%I64d", &num);
34             if(judge(num)==true){
35                 ans++;
36             }
37         }
38         printf("%d
", ans);
39     }
40     return 0;
41 }

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Problem B The Fibonacci Segment

找出长度最长的斐波那契数。主要考虑n=1的特殊情况后扫描一遍。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <cstdlib>
 6 #include <algorithm>
 7 #define LEN 100010
 8 
 9 using namespace std;
10 
11 int p[LEN];
12 
13 int main()
14 {
15 //    freopen("in.txt", "r", stdin);
16     int n, ans;
17     while(scanf("%d", &n)!=EOF)
18     {
19         for(int i=0; i<n; i++){
20             scanf("%d", &p[i]);
21         }
22         if(n==1){
23             ans = 1;
24         }else{
25             ans = 2;
26             int loc = 2;
27             for(int i=2; i<n; i++){
28                 if(p[i] == p[i-1]+p[i-2]){
29                     loc++;
30                 }
31                 else{
32                     if(loc>ans)ans = loc;
33                     loc = 2;
34                 }
35             }
36             if(loc>ans)ans = loc;
37         }
38         printf("%d
", ans);
39     }
40     return 0;
41 }

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比较水只做出两道题。继续努力！

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