ACdream 简单数论 专题
A - ACdream运动会
分析: 暴力hash
代码:
//
// Created by TaoSama on 2015-09-24
// 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, m, a[25];
bool vis[N];
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i) scanf("%d", a + i);
int ans = 0;
memset(vis, false, sizeof vis);
for(int i = 1; i <= m; ++i) {
for(int j = a[i]; j <= n; j += a[i])
if(!vis[j]) ++ans, vis[j] = true;
}
printf("%d\n", ans);
}
return 0;
}
B - 数论基础知识测试
代码:
//
// Created by TaoSama on 2015-09-24
// 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 euler(int n) {
int ret = n, t = n;
for(int i = 2; i * i <= n; i++) {
if(t % i == 0) {
ret = ret / i * (i - 1);
while(t % i == 0) t /= i;
}
}
if(t > 1) ret = ret / t * (t - 1);
return ret;
}
int exgcd(int a, int b, int &x, int& y) {
int d = a;
if(!b) x = 1, y = 0;
else {
d = exgcd(b, a % b, y, x);
y -= (a / b) * x;
}
return d;
}
int inv(int a, int m) {
int x, y;
exgcd(a, m, x, y);
return (m + x % m) % m;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int op;
while(scanf("%d", &op) == 1) {
int ans;
if(op == 1) {
int n; scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
int x; scanf("%d", &x);
if(i == 1) ans = x;
else ans = __gcd(ans, x);
}
} else if(op == 2) {
int x, m; scanf("%d%d", &x, &m);
ans = inv(x, m);
} else {
int x; scanf("%d", &x);
ans = euler(x);
}
printf("%d\n", ans);
}
return 0;
}
C - 小Y喜欢拆拆拆
分析: 模拟
代码:
//
// Created by TaoSama on 2015-09-24
// 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;
bool isPrime(int x) {
for(int i = 2; i * i <= x; ++i)
if(x % i == 0) return false;
return true;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d", &n) == 1) {
set<int> s;
while(true) {
if(s.count(n)) {
n = -1;
break;
}
s.insert(n);
if(isPrime(n)) break;
else {
int t = n, sum = 0;
for(int i = 2; i * i <= n; ++i) {
while(t % i == 0) {
sum += i;
t /= i;
}
if(t == 1) break;
}
if(t > 1) sum += t;
n = sum;
}
}
printf("%d\n", n);
}
return 0;
}
D - NT的数字
分析: 被8整除的特点后三位能被整除, 特判 长度 < 3 的全排列 然后暴力枚举1000以内能被8整除的数字 看给出的能不能构造
代码:
//
// Created by TaoSama on 2015-09-24
// 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;
char s[1005];
int n, have[10];
bool judge() {
bool can = false;
for(int i = 0; i < 1000; ++i) {
if(i % 8) continue;
int d = 0, t = i, cur[10] = {};
while(t) {
cur[t % 10]++;
t /= 10;
d++;
}
if(d == 0) { //000
if(have[0] >= 3) can = true;
} else {
if(d == 1) cur[0] += 2;
if(d == 2) cur[0] += 1;
bool ok = true;
for(int j = 0; j < 10; ++j)
if(have[j] < cur[j]) ok = false;
can = ok;
}
if(can) break;
}
return can;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%s", s) == 1) {
memset(have, 0, sizeof have);
for(int i = 0; s[i]; ++i)
have[s[i] - '0']++;
n = strlen(s);
if(n <= 3) {
sort(s, s + n);
bool ok = false;
do {
int x = atoi(s);
if(x % 8 == 0) {
ok = true;
break;
}
} while(next_permutation(s, s + n));
puts(ok ? "YES" : "NO");
continue;
}
puts(judge() ? "YES" : "NO");
}
return 0;
}
E - NT的表白
题意: 求1111....1111%b = ?
分析1: 结论直接秒 b | a的前提下
111…111 % b= ((10^a - 1) / 9) % b = (10^a -1) % (9*b) / 9
详情看: 点击打开链接
分析2: 我的思路 根据 a = x * y => a % m = (x % m * y % m) %m 然后折半分治 O(logn)
代码:
//
// Created by TaoSama on 2015-09-24
// 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;
typedef long long LL;
LL x, m;
LL ksm(LL x, LL n, LL m) {
LL ret = 1;
while(n) {
if(n & 1) ret = ret * x % m;
x = x * x % m;
n >>= 1;
}
return ret;
}
LL dfs(LL n) {
if(n <= 5) {
LL x = 0;
for(int i = 1; i <= n; ++i)
x = (x * 10 + 1) % m;
return x;
}
LL tmp = dfs(n >> 1), mul = n & 1 ? 10 : 1;
// cout << "tmp: " << tmp << " mul: " << mul << " " << ksm(10, n >> 1, m) << endl;
return (tmp * ksm(10, n >> 1, m) % m * mul % m + tmp * mul % m + mul / 10) % m;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
// ios_base::sync_with_stdio(0);
while(scanf("%lld%lld", &x, &m) == 2) {
printf("%lld\n", dfs(x));
}
return 0;
}
分析3: f[n] = f[n-1]*10 + 1 然后构造矩阵 矩阵快速幂 同样是O(logn) 思路来自学长
代码: 就不贴了 很好写
F - 这题占坑备用
分析: 直接构造所有的NT数 注意也许可能爆LL? 转double判断一下 然后sort直接找就好了
代码:
//
// Created by TaoSama on 2015-09-24
// 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;
typedef long long LL;
vector<LL> all;
LL l, r;
int getLen(LL x) {
int ret = 0;
for(; x; x /= 10) ++ret;
return ret;
}
void dfs(LL x) {
for(int i = 2; i < 20; ++i) {
if(1.0 * x * i > 1e18) break;
LL y = x * i;
if(i == getLen(y)) {
all.push_back(y);
dfs(y);
}
}
}
void gao() {
for(int i = 0; i < 10; ++i) {
all.push_back(i);
if(i >= 5) dfs(i);
}
sort(all.begin(), all.end());
// for(auto v : all) printf("%lld ", v); puts("");
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
gao();
while(scanf("%lld%lld", &l, &r) == 2) {
int ans = upper_bound(all.begin(), all.end(), r) -
lower_bound(all.begin(), all.end(), l);
printf("%d\n", ans);
}
return 0;
}