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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 .
For example : There are 5 soldiers (A B C D E)with level numbers : 2 4 5 6 4; One method : C could teach B; B could teach A; So , A B C are eligible to study on the same broomstick. D could teach E;So D E are eligible to study on the same broomstick; Using this method , we need 2 broomsticks. Another method: D could teach A; So A D are eligible to study on the same broomstick. C could teach B; So B C are eligible to study on the same broomstick. E with no teacher or student are eligible to study on one broomstick. Using the method ,we need 3 broomsticks. ……After checking up all possible method, we found that 2 is the minimum number of broomsticks needed.
Input
Input file contains multiple test cases.
In a test case,the first line contains a single positive number N indicating the number of soldiers.(0<=N<=3000) Next N lines :There is only one nonnegative integer on each line , indicating the level number for each soldier.( less than 30 digits);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 4Sample Output
1
2主要问题：输入的数过大，需要用大数。输入小于1030，输入只能用字符串，但保存可以用两个long long数据来存，因此，可以设置一个结构体num{long long a,b;}，a存放高位，b存放低位。
突破口：将数据从大到小排序，出现次数（重复次数）最多的那个数，其出现的次数就是结果。代码如下：
#include#include #include #include using namespace std;struct num{ long long a; long long b;}m[3005];bool cmp(num x,num y) //结构体是个好东西{ if(x.a!=y.a) return x.a>y.a; else return x.b>y.b;}int main(){ int n; int i,j,k,p,q,len,max,temp; char s[32],c; while(~scanf("%d",&n)) { c=getchar(); //要用c吃掉一个过行符 max=temp=1; for(i=0;i<=n-1;i++) //输入并处理数据 { scanf("%s",s); m[i].a=m[i].b=0; len=strlen(s)-1; for(k=p=q=0,j=len;j>=0;j--) if(k<=16) //一个long long存放16位 { m[i].b+=(s[j]-'0')*(long long)pow(10,p++); k++; } else { m[i].a+=(s[j]-'0')*(long long)pow(10,q++); } } sort(m,m+n,cmp); // for(i=0;i<=n-1;i++) // printf("%016lld %016lld\n",m[i].a,m[i].b); if(n==0) printf("0\n"); else { for(i=1;i<=n-1;i++) { if(m[i].a==m[i-1].a&&m[i].b==m[i-1].b) { temp++; } else { if(temp>max) max=temp; temp=1; } } if(temp>max) //最后出来还要再比较一便，以防出现意外情况（即所有数都相等） max=temp; printf("%d\n",max); } } return 0;}
分析： 与那道木棍的题很相似，不过这道题算法相对更简单一点，除了需要用到大数比较复杂以外，其它不会很难。
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