源地址:http://poj.org/problem?id=1840
用下map就行。我们先处理前两个数,将它们的结果放入map中,然后暴力枚举后三个的结果,看看map中有没有它的相反数即可。
不过这里好像得有一个比较优美的姿势才行。刚开始算一个结果后,都是直接mp[res]++ 或者 ans += mp[res],这样子好像会超时。对于每得到一个数,我们最好先用map.find() 与map.end()比较下,如果想等,那么说明不存在这个数。这样能快很多。新节能 get()√
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
int n,m,t;
#define Mod 1000000007
#define N 110
#define M 1000100
map<int,int> mp;
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#endif
int a,b,c,d,e;
while(scanf("%d%d%d%d%d",&a,&b,&c,&d,&e)!=EOF){
mp.clear();
for(int i=-50;i<=50;i++){
for(int j=-50;j<=50;j++){
if(i == 0 || j == 0) continue;
int tmp = a*i*i*i+b*j*j*j;
if(mp.find(tmp) == mp.end())<span style="white-space:pre"> </span>//
mp[tmp] = 1;
else
mp[tmp]++;
}
}
int ans = 0;
for(int i=-50;i<=50;i++){
for(int j=-50;j<=50;j++){
for(int k=-50;k<=50;k++){
if(i == 0 || j == 0 || k == 0) continue;
int tmp = c*i*i*i+d*j*j*j+e*k*k*k;
if(mp.find(0-tmp) == mp.end())<span style="white-space:pre"> </span>//
continue;
ans += mp[(0-tmp)];
}
}
}
printf("%d\n",ans);
}
return 0;
}