求1~n與x互質的數的個數（6個題、容斥原理）
阿新 • • 發佈：2019-01-22
HDU 4135、POJ 2773、HDU 1695、HDU 2841、ZOJ 2836、HDU 1796
HDU 4135 Co-prime
題意: 求
分析: 裸題
程式碼:
//
// Created by TaoSama on 2015-10-24
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath> POJ 2773 Happy 2006
題意: 求第
分析: 裸題上二分一下
程式碼:
//
// Created by TaoSama on 2015-10-24
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, k;
typedef long long LL;
vector<int> factor;
void fact(int x) {
factor.clear();
for(int i = 2; i * i <= x; ++i) {
if(x % i == 0) {
factor.push_back(i);
while(x % i == 0) x /= i;
}
if(x == 1) break;
}
if(x > 1) factor.push_back(x);
}
LL calc(LL x) {
LL ret = 0;
int sz = factor.size();
for(int s = 1; s < 1 << sz; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < sz; ++i)
if(s >> i & 1) ++cnt, mul *= factor[i];
if(cnt & 1) ret += x / mul;
else ret -= x / mul;
}
return x - ret;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d", &n, &k) == 2) {
fact(n);
LL l = 1, r = 1e10;
while(l <= r) {
LL m = l + r >> 1;
if(calc(m) < k) l = m + 1;
else r = m - 1;
}
printf("%I64d\n", l);
}
return 0;
}HDU 1695 GCD
題意:
分析:
程式碼:
//
// Created by TaoSama on 2015-10-24
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
vector<int> factor[N];
void gao() {
bool vis[N] = {};
for(int i = 2; i < N; ++i) {
if(vis[i]) continue;
for(int j = i; j < N; j += i) {
vis[j] = true;
factor[j].push_back(i);
}
}
}
typedef long long LL;
int calc(int k, int x) {
int ret = 0;
int sz = factor[k].size();
for(int s = 1; s < 1 << sz; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < sz; ++i) {
if(s >> i & 1) ++cnt, mul *= factor[k][i];
}
if(cnt & 1) ret += x / mul;
else ret -= x / mul;
}
return x - ret;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
int kase = 0;
gao();
while(t--) {
int l, r, k; scanf("%d%d%d%d%d", &l, &l, &r, &r, &k);
if(k == 0) {printf("Case %d: 0\n", ++kase); continue;}
l /= k, r /= k;
if(l > r) swap(l, r);
LL ans = 0;
for(int i = 1; i <= l; ++i)
ans += calc(i, r) - calc(i, i - 1);
printf("Case %d: %I64d\n", ++kase, ans);
}
return 0;
}HDU 2841 Visible Trees
題意:
分析:
程式碼:
//
// Created by TaoSama on 2015-10-24
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
vector<int> factor[N];
void gao() {
bool vis[N] = {};
for(int i = 2; i < N; ++i) {
if(vis[i]) continue;
for(int j = i; j < N; j += i) {
vis[j] = true;
factor[j].push_back(i);
}
}
}
int calc(int k, int x) {
int ret = 0;
int sz = factor[k].size();
for(int s = 1; s < 1 << sz; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < sz; ++i)
if(s >> i & 1) ++cnt, mul *= factor[k][i];
if(cnt & 1) ret += x / mul;
else ret -= x / mul;
}
return x - ret;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
gao();
while(t--) {
int n, m; scanf("%d%d", &n, &m);
if(n > m) swap(n, m);
long long ans = 0;
for(int i = 1; i <= n; ++i) ans += calc(i, m);
printf("%I64d\n", ans);
}
return 0;
}ZOJ 2836 Number Puzzle
題意:
分析:
程式碼:
//
// Created by TaoSama on 2015-10-24
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;