| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 16095 | Accepted: 7902 |
Description
Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}. Determine how many solutions satisfy the given equation.Input
The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.Output
The output will contain on the first line the number of the solutions for the given equation.Sample Input
37 29 41 43 47Sample Output
654Source
Romania OI 2002題目意思：
有一個五元一次方程，給出五個系數，解的范圍是[-50,0)(0,50]，求解的個數。解題思路：
直接五層for循環肯定就爆了，所以原式變形成：-(a1x1^3+ a2x2^3)=a3x3^3+ a4x4^3+ a5x5^3=0。哈希數組的下標表示求出的數，數組的值表示這個數出現的次數（類似于桶排序的思想）。①先求出等式左邊的值，將對應下標哈希數組的值加1，因為2*50*50^4=25000000，所以最多有25000000種不同的值。注意這個值可能為負，所以負數時要將其加上25000000再進行處理。②然后枚舉出等式右邊的值，判斷是否在出現過這個值，然后利用哈希數組將解的個數加上這個數出現的次數。同樣地這個值可能為負，所以負數時要將其加上25000000再進行處理。#include<iostream>#include<cstdio>#include<iomanip>#include<cmath>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;/*a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0*/short ha[25000000];//左半部分的值int main(){#ifdef ONLINE_iUDGE#else //freopen("F:/cb/read.txt","r",stdin); //freopen("F:/cb/out.txt","w",stdout);#endif ios::sync_with_stdio(false); cin.tie(0); memset(ha,0,sizeof(ha)); int a1,a2,a3,a4,a5; //系數 while(cin>>a1>>a2>>a3>>a4>>a5) { int a,b,c,d,e,ans=0; for(a=-50; a<=50; ++a) for(b=-50; b<=50; ++b) if(a!=0&&b!=0) { int t=a*a*a*a1+b*b*b*a2; t=-t; if(t<0) t+=25000000;//負值 ++ha[t]; } for(c=-50; c<=50; ++c) for(d=-50; d<=50; ++d) for(e=-50; e<=50; ++e) if(c!=0&&d!=0&&e!=0) { int num=c*c*c*a3+d*d*d*a4+e*e*e*a5; if(num<0) num+=25000000; ans+=ha[num]; } cout<<ans<<endl; } return 0;}/*37 29 41 43 47*/
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