ZOJ Monthly, May 2008 部分题目解题报告

Source: ZOJ 2956 - 2964

http://acm.zju.edu.cn/onlinejudge/showProblems.do?contestId=1&pageNumber=20

ZOJ 2956 Another Horizontally Visible Segments

题意：给一些平行y轴的线段，端点都是integer。平行x轴地切一刀，问最多能切多少线段。

分析：坐标都是整点而且范围很小，所以就线段扫描一下就可以。o(n)

#include<cstdio>
#include<cstring>
using namespace std;

const int ps = 100001;
int mpf[ps+5], prime[ps+5];
int pn = 0;
int T, a, z;

void getprime(){
memset(mpf, 0, sizeof(mpf));
for (int i = 2; i < ps; i ++){
if (mpf[i] == 0) prime[pn++] = i;
for (int j = 0; j < pn && i*prime[j] < ps && (prime[j] <= mpf[i]||mpf[i] == 0); j++)
mpf[i*prime[j]] = prime[j];
}
}
int phi(int x){
int ret = 1;
for (int i = 0; i < pn; i++){
if (x % prime[i] == 0){
int tmp = prime[i] - 1;
x /= prime[i];
while(x % prime[i] == 0){
tmp *= prime[i];
x /= prime[i];
}
ret *= tmp;
}
}
if (x > 1){
ret *= x - 1;
}
return ret;
}
int pow_mod(int a, int n){
if (n == 0) return 1;
if (n == 1) return a % z;
int tmp = pow_mod(a, n/2);
if (n & 1) return (long long) tmp * tmp % z * a % z;
return (long long) tmp * tmp % z;
}
int cal(int a, int x){
int ans = x;
for (int i = 1; i * i <= x; i++){
if (x % i == 0){
int tmp = pow_mod(a, i);
if (tmp == 1) return i;
tmp = pow_mod(a, x/i);
if (tmp == 1) ans = x/i;
}
}
return ans;
}
int main()
{
getprime();
scanf("%d", &T);
while(T--)
{
scanf("%d %d", &a, &z);
int ph = phi(z);
ph = cal(a, ph);
printf("%lld\n", (long long)ph * 6 + 3);
}
return 0;
}

View Code