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题意:给定一个四元二次方程的系数a,b,c,d;问有多少个解；
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}.
题目没有说清楚当a = b时，是否x1与x2可以看成一个即之后是乘以16还是要根据情况相乘；出题者意思是不考虑这些
思路:直接枚举两个x1,x2的可能的和值，在另外两个里面找出相反数的个数即可；用STL来查找重复元素的个数即可；
时间复杂度为O(n^2log(n^2)) n <= 100;
#include<iostream> #include<cstdio> #include<cstring> #include<string.h> #include<algorithm> #include<vector> #include<cmath> #include<stdlib.h> #include<time.h> #include<stack> #include<set> #include<map> #include<queue> using namespace std; #define rep0(i,l,r) for(int i = (l);i < (r);i++) #define rep1(i,l,r) for(int i = (l);i <= (r);i++) #define rep_0(i,r,l) for(int i = (r);i > (l);i--) #define rep_1(i,r,l) for(int i = (r);i >= (l);i--) #define MS0(a) memset(a,0,sizeof(a)) #define MS1(a) memset(a,-1,sizeof(a)) #define MSi(a) memset(a,0x3f,sizeof(a)) #define inf 0x3f3f3f3f #define lson l, m, rt << 1 #define rson m+1, r, rt << 1|1 typedef __int64 ll; template<typename T> void read1(T &m) { T x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} m = x*f; } template<typename T> void read2(T &a,T &b){read1(a);read1(b);} template<typename T> void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);} template<typename T> void out(T a) { if(a>9) out(a/10); putchar(a%10+‘0‘); } int A[2][10010],top,f[110]; int solve(int n,int m,int id) { top = 0; rep1(i,1,100){ int t = n*f[i]; rep1(j,1,100) A[id][top++] = t + m*f[j]; } sort(A[id],A[id] + top); } int main() { int a,b,c,d; rep1(i,1,100) f[i] = i*i; while(scanf("%d%d%d%d",&a,&b,&c,&d) == 4){ if((a > 0 && b > 0 && c > 0 && d > 0) ||(a < 0 && b < 0 && c < 0 && d < 0)) out(0); else{ ll ans = 0; solve(a,b,0);solve(c,d,1); rep0(i,0,top){ ans += (upper_bound(A[1],A[1]+top,-A[0][i]) - lower_bound(A[1],A[1]+top,-A[0][i])); } out(ans*16); } puts(""); } return 0; }
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原文地址：http://www.cnblogs.com/hxer/p/5225816.html
