zoj 2839 Find the Sequences（数学题）

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题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1839

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题意：

要找所有满足条件的等差数列的首项和等差！

条件是等差数列的每一项（共n项），都能用小于等于m的两个数字的三次方的和（i*i*i+j*j*j == num）；

要按等差从小到大输出，如果等差相同就按首项由小到大！

Find the Sequences

Time Limit: 2 Seconds Memory Limit: 65536 KB

The sequence we discuss here is in the form a, a + b, a + 2b, a + 3b, ... where a is a non-negative integer and b is a positive integer. Your task is to find out all such sequences whose
elements are of the form p3+q3 where p and q are non-negative integers.
Input:
The input consists of multiple test cases.
In each case, there're two integers n, m (3 <= n <= 10, 1 <= m <= 50). n, as mentioned above, is the number of terms in sequence, and m is the upper bound of p and q, which means 0 <=
p, q <=m.
The input will be ended with "0 0" which should not be processed.
Output:
For each test case, output "Case #:" first. "#" is the number of the case, which starts from 1.
If no such sequence was found, simply output "NONE". Otherwise output the sequences in multiple lines, each line contains two integers a and b mentioned above. The sequence with smaller
b comes first, if there's a tie, the one with smaller a comes first.
Separate two consecutive test cases with a blank line, but Do NOT output an extra blank line after the last one.
Sample Input:

4 10
9 1
0 0

Sample Output:

Case 1:
152 64
27 162

Case 2:
NONE

Author: DAI, Wenbin

Source: Zhejiang University Local Contest 2007
思路就是：先算出所有小于等于m的数字的所有三次的可能；

等差数列的每一下必然就在算出的这些数字中，这样大大可以减少时间；

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct node
{
	int a;
	int b;
}no[5000];
bool cmp(node e,node y)
{
	if(e.b==y.b)
		return e.a<y.a;
	return e.b<y.b;
}
int r[5000],f[300000];
int main()
{
	int n,m,p,q,i,j,k,KK,a1,a2,bb,x,count,flag,ans,u,pp,a3,yy;
	KK=1;
	while(~scanf("%d%d",&n,&m)&&(n+m))
	{
		if(KK!=1)
			printf("\n");
		printf("Case %d:\n",KK);
		KK++;
		k=0;
		memset(f,0,sizeof(f));
		for(i=0;i<=m;i++)
		{
			for(j=0;j<=m;j++)
			{
				pp=i*i*i+j*j*j;
				if(f[pp]==0)
				{
					f[pp]=1;
					r[k++]=pp;
				}
			}
		}
		sort(r,r+k);
		flag=0;
		ans=0;
		u=0;
		for(i=0;i<k-n;i++)
		{
			a1=r[i];
			for(j=i+1;j<k;j++)
			{
				a2=r[j];
				bb=a2-a1;
				for(count=3;count<=n;count++)
				{
					a3=a1+(count-1)*bb;
					if(a3>r[k-1]||f[a3]==0)
						break;
				}
				if(count>n)
				{
					flag=1;
					no[u].a=a1;
					no[u++].b=bb;
				}
			}
		}
		sort(no,no+u,cmp);
		if(flag)
		{
			for(i=0;i<u;i++)
			{
				printf("%d %d\n",no[i].a,no[i].b);
			}
		}
		else
			printf("NONE\n");
	}
	return 0;
}