PAT Advanced 1041 Be Unique (20) [Hash散列]
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题目
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, 10^4]. 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
题目分析
- 定义数组ns[N],存放输入数字
- 定义数组ts[10001],记录输入数字出现次数
- 大小为10001,(题目已知:输入数字取值范围[1, 10^4])
- 数组下标--输入数字
- 数组元素--输入数字出现次数
- 遍历输入数字(ns数组),最早出现次数==1(表明是唯一数),打印退出
易错点
- 若没有唯一数时,打印None,注意None后不需要加" "就可以AC
Code
Code 01
#include <iostream>
//#include <>
using namespace std;
int main(int argc, char * argv[]){
int N,m;
scanf("%d",&N);
int ns[N];
int ts[10001]={0};
for(int i=0;i<N;i++){
scanf("%d",&ns[i]);
ts[ns[i]]++;
}
bool flag = false;
for(int i=0;i<N;i++){
if(ts[ns[i]]==1){
printf("%d",ns[i]);
flag = true;
break;
}
}
if(!flag)printf("None");
return 0;
}
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PAT (Advanced Level) 1041. Be Unique (20)