HDU 5847 构造

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Different Sums

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Special Judge

Problem Description

A subsum of the sequence is sum of one or more consecutive integers of it. You are given an integer

1≤N≤2000 ). Your task is to make a sequence of integers which are less than

3(N+6), such that its all subsums (

N(N+1)/2 in total) are different from each other.

Input

There are several test cases.

The first line of the input contains an integer

T(1≤T≤200), the number of test cases.

Each of the next

T lines contains an integer ,

N the length of the sequence.

Output

For each test case, print one line with

N space separated integers representing your sequence.

If multiple solutions exist, any of them will be accepted.

Sample Input

Sample Output

1 21 2 4 8 16
题意：给你一个n，找出一个长度为n的序列，数都为正整数且不超过3*n+18，使得他们的子序列和各不相同。
题解：我是个翻译员23333，具体为什么这样做我也不会QAQ。
翻译：寻找一个素数p刚好大于n，定义s[i]为前i个数的和，再找出一个x，0<=x<p
S[i] = 2 * i * p + (i * (i+1) / 2 * x) % p
定义r[i] = (i * (i+1) / 2 * x) % p.
如果s[i]-s[j]=s[k]-s[l]  那么i-j=k-l 因为|r[i]-r[j]|<p且|r[k]-r[l]|<p
而且(r[i]-r[j]-r[k]+r[l])能被p整除，所以i+j=k+l。
也就是说i=k且j=l
对于所有的6<=n<=2000，我们只要for一下寻找满足每个数都小于3*n+18的x就可以了
然后我们从1for到p-1就行了，不用去管0
ps：朝鲜ACM  best ACM  强无敌  这题全场744发没一个队过
#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int s[2005];int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int i,j,p;for(p=n+1;;p++){int flag=1;for(j=2;j*j<=p;j++){if(p%j==0){flag=0;break;}}if(flag)break;}int x;for(x=1;x<p;x++){for(i=1;i<=n;i++){s[i]=2*i*p+(i*(i+1)/2*x)%p;if(s[i]-s[i-1]>=3*n+18)break;}if(i==n+1)break;} for(i=1;i<n;i++){printf("%d ",s[i]-s[i-1]);}printf("%d\n",s[i]-s[i-1]);}return 0;}

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