Luogp1586 Sifang theorem and P1586 Sifang Theorem

Luogp1586 Sifang theorem and P1586 Sifang Theorem
Description

The square theorem is well known: any positive integer nn can be decomposed into a sum of squares of up to four integers. Example: 25 = 1 ^ {2} + 2 ^ {2} + 2 ^ {2} + 4 ^ {2} 25 = 12 + 22 + 22 + 42, of course there are other decomposition solutions, 25 = 4 ^ {2} + 3 ^ {2} 25 = 42 + 32 and 25 = 5 ^ {2} 25 = 52. The total number of solutions that can be decomposed by the given positive integer nn. Note: 25 = 4 ^ {2} + 3 ^ {2} 25 = 42 + 32 and 25 = 3 ^ {2} + 4 ^ {2} 25 = 32 + 42 solution.

Input/Output Format

Input Format:

 

The first behavior is a positive integer tt (t \ le 100t ≤ 100), and the next tt row contains a positive integer nn (n \ le 32768n ≤ 32768 ).

 

Output Format:

 

For each positive integer nn, the total number of output solutions.

 

Input and Output sample input sample #1: Copy

12003

Output sample #1: Copy 48

$ N ^ 4 $ the positive solution of brute force is a backpack $ dp [I] [j] $, which indicates the number of solutions $ j $

// luogu-judger-enable-o2#include<iostream>#include<cstdio>#define LL long long using namespace std;const int MAXN=1e5+10;int dp[5][MAXN];int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    dp[0][0]=1;    for(register int i=1;i<=200;i++)        for(register int j=1;j<=4;j++)            for(register int k=1;k<=32768;k++)                if(i*i<=k)                    dp[j][k]+=dp[j-1][k-i*i];    int T;        scanf("%d",&T);    while(T--)    {        int a;        scanf("%d",&a);        printf("%d\n",dp[1][a]+dp[2][a]+dp[3][a]+dp[4][a]);    }    return 0;}

 

// luogu-judger-enable-o2#include<iostream>#include<cstdio>#define LL long long using namespace std;const int MAXN=1e6+10;int mul[MAXN],dp[MAXN];int ans[MAXN];int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    int N=200;    for(int i=1;i<=N;i++) mul[i]=i*i;    for(int i=1;i<=N;i++) ans[ mul[i] ] ++;    for(int i=1;i<=N;i++)        for(int j=i;j<=N;j++)            ans[ mul[i]+mul[j] ] ++;    for(int i=1;i<=N;i++)        for(int j=i;j<=N;j++)            for(int k=j;k<=N;k++)                ans[ mul[i]+mul[j]+mul[k] ] ++;    for(int i=1;i<=N;i++)        for(int j=i;j<=N;j++)            for(int k=j;k<=N;k++)                for(int l=k;l<=N;l++)                    ans[ mul[i]+mul[j]+mul[k]+mul[l] ]++;    int T;    scanf("%d",&T);    while(T--)    {        int a;        scanf("%d",&a);        printf("%d\n",ans[a]);    }        return 0;}