题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1041
题目描述:
Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1, 104]. The first one who bets on a unique number wins. For example, if there are 7 people betting on 5 31 5 88 67 88 17, then the second one who bets on 31 wins.
Input Specification:
Each input file contains one test case. Each case contains a line which begins with a positive integer N (<=105) and then followed by N bets. The numbers are separated by a space.
Output Specification:
For each test case, print the winning number in a line. If there is no winner, print "None" instead.
Sample Input 1:7 5 31 5 88 67 88 17Sample Output 1:
31Sample Input 2:
5 888 666 666 888 888Sample Output 2:
None
分析:(1)直接用查询表。(2)有个坑爹的陷阱,用cin输入会超时,但是用scanf输入去可以通过。原因是???
参考代码:
#include<iostream> #include<cstdio> #include<string.h> using namespace std; int result[10000 + 5]; int input[100000 + 5]; int main() { int n; cin>>n; int i; int temp; bool flag = false; for(i=0; i<n; i++) { //cin>>input[i]; 用cin会超时, 用scanf不会 scanf("%d",&input[i]); result[ input[i] ]++; } for(i=0; i<n; i++) { if(result[ input[i] ] == 1) { flag = true; break; } } if(flag) cout<<input[i]<<endl; else cout<<"None"<<endl; return 0; }