二分寻找解的个数
1.HDU1496 Equations
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.
End of file.
1 2 3 -4 1 1 1 1
39088 0
先把前两个数的和的所有方案找出 最后二分寻找最后两个数的和
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 10010;
int k,a,b,c,d;
long long ans;
int s[maxn];
void init(){
k=0;
for(int i=1;i<=100;i++)
for(int j=1;j<=100;j++)
s[k++]=i*i*a+b*j*j;
}
void binary_search(int t){
int l=0,r=k;
while(l<r){
int mid=(l+r)/2;
if(s[mid]==t){
ans++;
for(int ss=mid-1;s[ss]==t&&ss>=1;ss--)
if(s[ss]==t)
ans++;
for(int ss=mid+1;s[ss]==t&&ss<k;ss++)
if(s[ss]==t)
ans++;
return;
}
if(t>s[mid])
l=mid+1;
else r=mid;
}
}
int main()
{
while(cin>>a>>b>>c>>d){
init();
ans = 0;
if(a>0&&b>0&&c>0&&d>0){
cout<<"0"<<endl;
continue;
}
sort(s,s+k);
for(int i=1;i<=100;i++){
for(int j=1;j<=100;j++){
int t=-(c*i*i+d*j*j);
int l=0,r=k;
binary_search(t);
}
}
printf("%I64d\n",ans*16);
}
return 0;
}
UVA1152 4 Values whose Sum is 0
Description
The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 228 ) that belong respectively to A, B, C and D .
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
For each input file, your program has to write the number quadruplets whose sum is zero.
Sample Input
1 6 -45 22 42 -16 -41 -27 56 30 -36 53 -37 77 -36 30 -75 -46 26 -38 -10 62 -32 -54 -6 45
Sample Output
5
Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[4010][4];
int num1[4001*4001];
int num2[4001*4001];
int main()
{
int cas,n;
scanf("%d",&cas);
while(cas--){
scanf("%d",&n);
for(int i=0;i<n;i++)
for(int j=0;j<4;j++)
scanf("%d",&a[i][j]);
int cnt=0;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
num1[cnt]=a[i][0]+a[j][1];
num2[cnt]=a[i][2]+a[j][3];
cnt++;
}
}
int count=0;
sort(num1,num1+cnt);
sort(num2,num2+cnt);
for(int i=0;i<cnt;i++){
int l=0,r=cnt-1;
while(l<=r)
{
int mid=(l+r)/2;
if(num1[i]+num2[mid]==0)
{
for(int j=mid;j>=0&&num2[j]==num2[mid];j--)
count++;
for(int j=mid+1;j<cnt&&num2[j]==num2[mid];j++)
count++;
break;
}
else if(num1[i]+num2[mid]<0)
l=mid+1;
else r=mid-1;
}
}
cout<<count<<endl;
if(cas!=0)
puts("");
}
return 0;
}