pku 3090 Visible Lattice Points

http://acm.pku.edu.cn/JudgeOnline/problem?id=3090

 

题目大意：一个从(0,0)到(n,n)的坐标阵，问多少点能被点(0,0)看到。如果(0,0)到(i,j)的连线被点挡住就算看不到。

画图可见，左上和右下两部分的对称性，考虑右下部分，as for a point(x,y)

1.有x>=y 

2.若gcd(x,y)>1;(比如(6,2))则t=gcd(x,y),必存在比它小的(x/t,y/t)让它穿过去(e.g(3,1))

那么我们求的就是<=num且与num互质的数的个数！=》欧拉函数

所以，ans(x)=2*sum(oula(y))+1;//y<=x,1是对角线上那个数

#include<iostream> #include<cmath> using namespace std; #define N 1005 int ans[N]; int pp[N]; int prime[168] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997}; int oula(int num) { int i; for(i=0;i<168&&prime[i]<=num;i++) if(prime[i]==num) return num-1; int re = num,tt=num; for(i=0;i<168&&prime[i]<=sqrt(tt*1.0)&&num!=1;i++) { bool flag=false; while(num%prime[i]==0) { num/=prime[i]; flag=true; } if(flag) re-=re/prime[i]; } if(num!=1) re-=re/num; return re; } int main() { int n,T,cnt; memset(ans,0,sizeof(ans)); ans[1]=1; for(int i=2;i<=N;i++) ans[i]=ans[i-1]+oula(i); //ans[i]=sum(oula(j)),j<=i; scanf("%d",&T); for(cnt=1;cnt<=T;cnt++) { scanf("%d",&n); printf("%d %d %d/n",cnt,n,ans[n]*2+1); } }