1041. Be Unique (20)【水题】——PAT (Advanced Level) Practise
2015-12-06 00:12
513 查看
题目信息
1041. Be Unique (20)时间限制100 ms
内存限制65536 kB
代码长度限制16000 B
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 (<=10^5) 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 17
Sample Output 1:
31
Sample Input 2:
5 888 666 666 888 888
Sample Output 2:
None
解题思路
记录每个数出现次数,第一个出现一次的为所求AC代码
#include <cstdio> #include <map> #include <vector> using namespace std; int main() { int n, t; vector<int> v; map<int, int> mp; scanf("%d", &n); for (int i = 0; i < n; ++i){ scanf("%d", &t); v.push_back(t); mp[t]++; } bool flag = false; for (int i = 0; i < v.size(); ++i){ if (mp[v[i]] == 1){ printf("%d\n", v[i]); flag = true; break; } } if (!flag){ printf("None\n"); } return 0; }
相关文章推荐
- libdvbpsi源码分析(四)PAT表解析/重建
- PAT配置
- 什么是端口复用动态地址转换(PAT) 介绍配置实例
- MikroTik layer7-protocol
- PAT是如何工作的
- PAT 乙级题:1002. 写出这个数 (20)
- PAT (Advanced Level) Practise 1001-1010
- 数据结构学习与实验指导(一)
- PAT Basic Level 1001-1010解题报告
- 1001. 害死人不偿命的(3n+1)猜想
- 1002. 写出这个数
- 1032. 挖掘机技术哪家强
- 1001. 害死人不偿命的(3n+1)猜想 (PAT basic)
- 1002. 写出这个数(PAT Basic)
- 1004. 成绩排名(PAT Basic)
- 1006. 换个格式输出整数(PAT Basic)
- 1007. 素数对猜想(PAT Basic)
- 1008. 数组元素循环右移问题
- 1009. 说反话(PAT Basic)
- 1011. A+B和C(PAT Basic)