数论 毕达哥斯拉三元组 + 欧拉函数 + 容斥原理 hdu3939

涉及的知识点还是挺多的，这题时间特卡。。。。

1.毕达哥斯拉三元组：

三元组（a，b，c),其中a，b，c无公因数，且满足a² +b² =c²。

a为奇数，b为偶数，c为奇数

可以得到如下勾股数组定理：

a = 2m*n;

b = m^2 - n^2;

c = m^2 + n^2;

其中m,n奇偶性不同

2.欧拉函数：

enlur[n]小于n且与n互素的数字个数

3.容斥原理

//毕达哥斯拉三元组，欧拉函数，容斥原理/********************************************Author         :CrystalCreated Time   :File Name      :********************************************/#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>#include <cstring>#include <climits>#include <string>#include <vector>#include <cmath>#include <stack>#include <queue>#include <set>#include <map>#include <sstream>#include <cctype>using namespace std;typedef long long ll;typedef pair<int ,int> pii;#define MEM(a,b) memset(a,b,sizeof a)#define CLR(a) memset(a,0,sizeof a);const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;#define maxn 1000005//#define LOCALll ans = 0;int euler[maxn];int s;int v[maxn];int primer[maxn];int gget[maxn];int x = 0;void init(){euler[1] = 1;for(int i=1;i<maxn;i++)euler[i] = i;for(int i=2;i<maxn;i++){if(euler[i] == i){primer[i] = 1;gget[x++] = i;for(int j=i;j<maxn;j+=i){euler[j] = euler[j]/i*(i-1);}}}}void check(int a){s = 0;if(primer[a]){v[s++] = a;return;}int res = a;for(int i=0;gget[i]*gget[i]<=a;i++){if(res%gget[i]==0){v[s++] = gget[i];while(res%gget[i]==0)res/=gget[i];}}if(res > 1)v[s++] = res;}void solve(int cnt, int state ,int pos,int b){if(pos == s){if(cnt%2){ans -= b/state;}else ans += b/state;return;}solve(cnt+1,state*v[pos],pos+1,b);solve(cnt,state,pos+1,b);}int main(){#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifinit();int t;cin >> t;while(t--){ans = 0;ll l;cin >> l;ll a = (ll)sqrt(l);for(ll i = a;i>=1;i--){ll b = (ll)sqrt(l-i*i);if(i%2==0){if(b>=i){ans += euler[i];}else{check(i);solve(0,1,0,b);}}else{check(i);if(b>=i){solve(0,1,0,i/2);}else solve(0,1,0,b/2);}}cout << ans << endl;}return 0;}