# UVA 11063 B2-Sequence
## 題目連結 [UVA 11063](https://vjudge.net/problem/UVA-11063)
### 題目內容
A B2-Sequence is a sequence of positive integers 1 ≤ b~1~ < b~2~ < b~3~ . . . such that all pairwise sums b~i~ +b~j~ , where i ≤ j, are different. Your task is to determine if a given sequence is a B2-Sequence or not.
### 輸入限制
Each test case starts with 2 ≤ N ≤ 100, the number of elements in a sequence. Next line will have N integers, representing the value of each element in the sequence. Each element b~i~ is an integer such that b~i~ ≤ 10000. There is a blank line after each test case. The input is terminated by end of file (EOF).
### 輸出限制
For each test case you must print the number of the test case, starting from 1, and a message indicating if the corresponding sequence it is a B2-Sequence or not. See the sample output below. After each test case you must print a blank line.
### 解題思路
1.要很注意題目的要求,每一項都不能低於1,數列要是遞增的,可以自己加自己
2.利用vector來存取每一項相加的內容,count來算次數,如果有超過1以上代表有重複
### 程式碼
```cpp=
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,kase=1;
while(cin>>n){
int a[n];
int ty=1;
vector<int> q;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]<1){
ty=0;
}
if(i!=0 && a[i]<=a[i-1]){
ty=0;
}
}
int x=0;
if(ty){
for(int i=0;i<n;i++){
for(int j=i;j<n;j++){
x=a[i]+a[j];
q.push_back(x);
if(count(q.begin(),q.end(),x)>1){
ty=0;
break;
}
}
if(ty==0){
break;
}
}
}
if(ty){
printf("Case #%d: It is a B2-Sequence.\n",kase++);
}
else{
printf("Case #%d: It is not a B2-Sequence.\n",kase++);
}
cout<<"\n";
}
}
```
## 測資
### Sample input
4
1 2 4 8
4
3 7 10 14
### Sample output
Case #1: It is a B2-Sequence.
Case #2: It is not a B2-Sequence.
## 中文題目連結 [zerojudge d123](https://zerojudge.tw/ShowProblem?problemid=d123)