HDU1800Flying to the Mars
http://acm.hdu.edu.cn/showproblem.php?pid=1800
輸出出現最多數字的個數
| In the year 8888, the Earth is ruled by the PPF Empire . As the population growing , PPF needs to find more land for the newborns . Finally , PPF decides to attack Kscinow who ruling the Mars . Here the problem comes! How can the soldiers reach the Mars ? PPF convokes his soldiers and asks for their suggestions . “Rush … ” one soldier answers. “Shut up ! Do I have to remind you that there isn’t any road to the Mars from here!” PPF replies. “Fly !” another answers. PPF smiles :“Clever guy ! Although we haven’t got wings , I can buy some magic broomsticks from HARRY POTTER to help you .” Now , it’s time to learn to fly on a broomstick ! we assume that one soldier has one level number indicating his degree. The soldier who has a higher level could teach the lower , that is to say the former’s level > the latter’s . But the lower can’t teach the higher. One soldier can have only one teacher at most , certainly , having no teacher is also legal. Similarly one soldier can have only one student at most while having no student is also possible. Teacher can teach his student on the same broomstick .Certainly , all the soldier must have practiced on the broomstick before they fly to the Mars! Magic broomstick is expensive !So , can you help PPF to calculate the minimum number of the broomstick needed .
Input Input file contains multiple test cases.
Output For each case, output the minimum number of broomsticks on a single line.
Sample Input 4 10 20 30 04 5 2 3 4 3 4
Sample Output 1 2 |
#include <iostream> #include <math.h> #include <cmath> #include <map> #include<cstdio> #define ll long long//能用long long 就別寫int!!!!!!!!!!!!! using namespace std; int max(ll x,ll y) { return x>=y?x:y; } int main() { ll n; while(scanf("%lld",&n)!=EOF)//注意這裡不能用cin,否則會超時！！！！！！！！ { map<ll,ll> maps; ll maxsfff=-1; for(ll i=1; i<=n; i++) { ll k; scanf("%lld",&k); maps[k]++; maxsfff=max(maxsfff,maps[k]); } printf("%lld\n",maxsfff); } }