HDU - 5166 - Missing number && 5167 - Fibonacci

Missing number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 430    Accepted Submission(s): 233

Problem Description

There is a permutation without two numbers in it, and now you know what numbers the permutation has. Please find the two numbers it lose.

 

Input

There is a number T shows there are T test cases below. (T≤10)
For each test case , the first line contains a integers n , which means the number of numbers the permutation has. In following a line , there are n distinct postive integers.(1≤n≤1,000)

 

Output

For each case output two numbers , small number first.

 

Sample Input

233 4 511

 

Sample Output

1 22 3

 

Source

BestCoder Round #28

 

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int vis[1005];int main() {int T, n;scanf("%d", &T);while(T--) {scanf("%d", &n);memset(vis, 0, sizeof(vis));int m = n + 2, a, b, flag = 1;while(n--) {scanf("%d", &a);vis[a] = 1;}for(int i=1; i<=m; i++) {if(!vis[i] && flag == 1) {a = i; flag = 2;}else if(!vis[i] && flag == 2) {b = i; break;}   }printf("%d %d\n", a, b);}return 0;} 

Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1312    Accepted Submission(s): 328

Problem Description

Following is the recursive definition of Fibonacci sequence:

Fi=⎧⎩⎨01Fi−1+Fi−2i = 0i = 1i > 1

Now we need to check whether a number can be expressed as the product of numbers in the Fibonacci sequence.

 

Input

There is a number T shows there are T test cases below. (T≤100,000)
For each test case , the first line contains a integers n , which means the number need to be checked. 
0≤n≤1,000,000,000

 

Output

For each case output "Yes" or "No".

 

Sample Input

3417233

 

Sample Output

YesNoYes

 

Source

BestCoder Round #28

AC代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <queue>#include <map>#define LL long longusing namespace std;const int MAX = 1000000010L;LL fibo[50] = {0, 1};map<LL, bool> ma;void init() {queue<LL> q;int flag;for(int i=2; i<50; i++) {fibo[i] = fibo[i-1] + fibo[i-2];if(fibo[i] > MAX) {flag = i; break;}}for(int i = 0; i < flag; i++) {q.push(fibo[i]); ma[fibo[i]] = true;}while(!q.empty()) {LL tmp = q.front();q.pop();for(int i = 2; i <= flag; i++) {LL t = tmp*fibo[i];if(t > MAX) break;if(ma[t]) continue;else {ma[t] = true;q.push(t);}}}}int main() {init();LL T, n;cin >> T;while(T--) {cin >> n;if(ma[n]) printf("Yes\n");else printf("No\n");} return 0;}