POJ 1840（哈希）

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题目链接：http://poj.org/problem?id=1840

看到这道题第一感觉是很熟悉，想起hdu上的一道题。

不过不同的是，这道题如果采用相同方法去做，数组要开始6kw，需要用short开，应该是数据不强吧。

另外，这道题map还可以过。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int INF=0x3f3f3f3f;const int maxn=62500010;const int maxnHash=31250000;int m,t,n;int a1,a2,a3,a4,a5;short hashTable[maxn];int p[50];int main(){#ifndef ONLINE_JUDGEfreopen("test.in","r",stdin);freopen("test.out","w",stdout);#endifwhile(~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5)){for(int i=-50;i<=50;i++){if(!i) continue;for(int j=-50;j<=50;j++){if(!j) continue;int tmp=a1*i*i*i+a2*j*j*j;hashTable[tmp+maxnHash]++;}}int ans=0;for(int i=-50;i<=50;i++){if(!i) continue;for(int j=-50;j<=50;j++){if(!j) continue;for(int k=-50;k<=50;k++){if(!k) continue;int tmp=a3*i*i*i+a4*j*j*j+a5*k*k*k;ans+=hashTable[-tmp+maxnHash];}}}printf("%d\n",ans);}return 0;}

附上hdu1496：

Consider equations having the following form:

a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.

It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.

Determine how many solutions satisfy the given equation.

Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
End of file.

Output
For each test case, output a single line containing the number of the solutions.

Sample Input
1 2 3 -4
1 1 1 1

Sample Output
39088
0

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int INF=0x3f3f3f3f;const int maxn=2000010;int a,b,c,d;int Hash[maxn];int p[105];void init(){    for(int i=1;i<=100;i++)        p[i]=i*i;}int main(){    init();    while(~scanf("%d%d%d%d",&a,&b,&c,&d)){        if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0)){            printf("0\n");            continue;        }        memset(Hash,0,sizeof(Hash));        for(int i=1;i<=100;i++){            for(int j=1;j<=100;j++){                Hash[(a*p[i]+b*p[j])+1000000]++;            }        }        int sum=0;        for(int i=1;i<=100;i++){            for(int j=1;j<=100;j++){                sum+=Hash[-(c*p[i]+d*p[j])+1000000];            }        }        printf("%d\n",sum*16);    }    return 0;}

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