Distribution money
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 511 Accepted Submission(s): 291
Problem Description
AFA want to distribution her money to somebody.She divide her money into n same parts.One who want to get the money can get more than one part.But if one man‘s money is more than the sum of all others‘.He shoule be punished.Each one
who get a part of money would write down his ID on that part.
Input
There are multiply cases.
For each case,there is a single integer n(1<=n<=1000) in first line.
In second line,there are n integer a1,a2...an(0<=ai<10000)ai is the the ith man‘s ID.
Output
Output ID of the man who should be punished.
If nobody should be punished,output -1.
Sample Input
3 1 1 2 4 2 1 4 3
Sample Output
1 -1
Source
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找出现数字大于2/n的。
#include <cstdio>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int main()
{
int n;
while(cin>>n)
{
int a[10005];
int i,x;
int ans;
memset(a,0,sizeof(a));
for(i=0; i<n; i++)
{
cin>>x;
a[x]++;
}
ans=n/2;
int flag=0;
for(i=0; i<=10000; i++)
{
if(a[i]>ans)
{
flag=1;
cout<<i<<endl;
break;
}
}
if(!flag)
cout<<-1<<endl;
}
return 0;
}
The mook jongTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 585 Accepted Submission(s): 419 Problem Description  ZJiaQ want to become a strong man, so he decided to play the mook jong。ZJiaQ want to put some mook jongs in his backyard. His backyard consist of n bricks that is 1*1,so it is 1*n。ZJiaQ want to put a mook jong in a brick. because of the hands of the mook jong， Input There ar multiply cases. For each case, there is a single integer n( 1 < = n < = 60) Output Print the ways in a single line for each case. Sample Input 1 2 3 4 5 6 Sample Output 1 2 3 5 8 12 Source Recommend hujie | We have carefully selected several similar problems for you: 5390 5389 5388 5387 5386 |
水dp。
#include <cstdio>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
ll m[111];
ll solve(int i)
{
if (m[i] != 0)
return m[i];
m[i] = solve(i-1) + solve(i-3) + 1;
return m[i];
}
int main()
{
int n;
m[0] = 0;
m[1] = 1;
m[2] = 2;
m[3] = 3;
while (cin >> n)
cout << solve(n) << endl;
return 0;
}
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