数论知识
LightOJ 1370 Bi-shoe and Phi-shoe
题意
找出比自己大且欧拉函数也比自己大的第一个数
思路
很巧妙, 找比自己大的第一个质数就可以， 因为质数x, ϕ ( x ) = x − 1 \phi(x) = x-1 ϕ(x)=x−1
代码
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 1e6 + 5;
int prime[maxn];
void doprime(){
memset(prime,0,sizeof(prime));
prime[1]=1;
for(int i=2;i*i<maxn;i++){
if(!prime[i]){
for(int j=2*i;j<maxn;j+=i){
prime[j]=1;
}
}
}
}
int solve(int n){
for(int i=n+1;;i++){
if(!prime[i])
return i;
}
}
int main(){
int t,n,num,cas=1;
doprime();
cin>>t;
while(t--){
long long ans = 0;
cin>>n;
for(int i=0;i<n;i++){
cin>>num;
ans += solve(num);
}
///Case 1:22
cout<<"Case "<<cas++<<": "<<ans<<" Xukha"<<endl;
}
}
LightOJ 1341 Aladdin and the Flying Carpet
题意
找出所有的约数对数, 并且满足对内的元素不小于b
思路
步骤：
- 筛质数
- 分解质因数，求约数个数
- 因为求对数ans，有重复，除以ans / 2，
- dfs筛所有约数
- 减去小于b的情况的约数
注意： 筛所有约数记得排序
代码
#include<iostream>
#include<cstring>
#include<algorithm>
#define dbg(a) cout<<#a<<" : "<<a<<endl
using namespace std;
typedef long long LL;
typedef pair<LL, int> PII;
const int N = 1000010;
int primes[N];
int cnt;
int cntf, cntd;
LL divisor[N];
PII factor[N];
bool st[N];
// 筛质数
void init() {
for(int i = 2; i < N; i ++){
if(!st[i]) primes[cnt ++] = i;
for(int j = 0; primes[j] * i < N; j ++) {
st[primes[j] * i] = true;
if(i % primes[j] == 0) break;
}
}
}
void dfs(LL u, LL p){
if(u > cntf) {
divisor[cntd ++] = p;
return;
}
for(int i = 0; i <= factor[u].second; i ++) {
dfs(u + 1, p);
p = (LL) p * factor[u].first;
}
}
int main(){
init();
int T;
cin >> T;
for(int k = 1; k <= T; k ++) {
cntd = cntf = 0;
LL a, b;
cin >> a >> b;
if(a< b * b) {
printf("Case %d: 0\n", k);
continue;
}
printf("Case %d: ", k);
LL sum = 1;
// 质因子分解
for(int i = 0; primes[i] <= a / primes[i]; i ++) {
LL p = primes[i];
if(a % p == 0) {
LL s = 0;
while(a % p == 0) {
a /= p;
s ++;
}
factor[++ cntf] = {p, s};
sum = (LL)sum * (s + 1);
}
}
if(a > 1) {
factor[++ cntf] = {a, 1};
sum = (LL)sum *(1 + 1);
}
sum /= 2;
//筛约数
dfs(1, 1);
sort(divisor, divisor + cntd);
for(int i = 0; i < cntd; i ++) {
if(b <= divisor[i]) break;
sum --;
}
printf("%lld\n", sum);
}
return 0;
}
LightOJ - 1282 Leading and Trailing
思路
太妙了，简直就是妙蛙种子吃着妙脆角妙进了米奇妙妙屋.
看大佬博客的。
- 取后三位，快速幂取模就行。LL trail = qmi(n, k, 1000);
- 取前三位就太喵了。
n k = 1 o l g n k = 1 0 k ∗ l g n n^k = 1o^{lgn^k} = 10^{k*lgn} nk=1olgnk=10k∗lgn
设 k ∗ l g n = a + b k*lg^n = a + b k∗lgn=a+b,a是值 k ∗ l g n k*lg^n k∗lgn 的整数部分, b是小数部分， 则 n k = 1 0 a ∗ 1 0 b n^k = 10^a * 10^b nk=10a∗10b。
可以看出 1 0 a 10^a 10a不会改变 n k n^k nk的大小，所以前三位取决于 1 0 b 10^b 10b
所以只需求b出来即可，怎么求呢?
用到个库函数可以求浮点数的小数部分,设为y。modf()
所以前三位为: pow(10, y) * 100;
若不足，输出语句添上补0即可
代码
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
LL qmi(LL a, LL b, LL p) {
LL res = 1;
while(b) {
if(b & 1) res = (LL)res * a % p;
a = (LL) a * a % p;
b >>= 1;
}
return res;
}
int main() {
int T;
cin >> T;
for(int i = 1; i <= T; i ++) {
LL n, k;
cin >> n >> k;
LL trail = qmi(n, k, 1000);
double x, y;
y = modf((double) k * log10(n), &x);
LL lead = pow(10, y) * 100;
printf("Case %d: %03lld %03lld\n",i, lead,trail);
}
return 0;
}
LightOJ - 1220 Mysterious Bacteria
思路
又是质因子分解咯
N
=
p
1
c
1
∗
p
2
c
2
∗
.
.
.
∗
p
k
c
k
N = p_1^{c1} * p_2^{c2} * ...* p_k^{c_k}
N=p1c1∗p2c2∗...∗pkck
要想化成
b
a
b^a
ba ，则
N
=
(
p
1
c
1
/
a
∗
p
2
c
2
/
a
∗
.
.
.
∗
p
k
c
k
/
a
)
a
N = (p_1^{c1/a} * p_2^{c2/a} * ...* p_k^{c_k/a})^a
N=(p1c1/a∗p2c2/a∗...∗pkck/a)a
则求
a
=
g
c
d
(
c
1
,
c
2
,
c
3
,
.
.
.
,
c
k
)
a = gcd(c1, c2, c3, ..., ck)
a=gcd(c1,c2,c3,...,ck)即可
注意当是负数的时候要特殊处理, 则幂
a
a
a 不能是偶数
代码
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100010;
int primes[N];
bool st[N];
int cnt;
void init() {
for(int i = 2; i < N; i ++){
if(!st[i]) primes[cnt ++] = i;
for(int j = 0; primes[j] * i < N; j ++) {
st[primes[j] * i] = true;
if(i % primes[j] == 0) break;
}
}
}
int gcd(int a, int b){
return b?gcd(b, a % b):a;
}
int main() {
init();
int T;
scanf("%d", &T);
for(int k = 1; k <= T; k ++) {
int n;
int f = 0;
scanf("%d", &n);
if(n < 0) {n = -n, f = 1;}
int d = 0;
for(int i = 0; i < cnt; i ++) { // primes[i] < n / primes[i]也可
if(n % primes[i] == 0) {
int s = 0;
while(n % primes[i] == 0) {s ++, n /= primes[i];}
if(d == 0) d = s;
else d = gcd(d, s);
}
}
if(n > 1) d = gcd(d, 1);
if(f) {
while(d % 2 == 0) d >>= 1;
}
printf("Case %d: %d\n",k, d);
}
return 0;
}
LightOJ - 1234 Harmonic Number
思路
结论题咯。
N较小，暴力求解。
N较大，
a
n
s
=
l
o
g
(
n
)
+
C
+
1.0
/
(
2
∗
n
)
,
(
C
=
0.57721566490153286060651209
)
欧
拉
常
数
ans = log(n) + C + 1.0/(2 * n), (C = 0.57721566490153286060651209) 欧拉常数
ans=log(n)+C+1.0/(2∗n),(C=0.57721566490153286060651209)欧拉常数
代码
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 100010;
const double r = 0.57721566490153286060651209;
double s[N];
int main() {
for(int i = 1; i < N; i ++) {
s[i] = s[i - 1] + 1.0 / i;
}
int T;
scanf("%d", &T);
for(int t = 1; t <= T; t ++) {
int n;
scanf("%d", &n);
if(n < N) printf("Case %d: %.10f\n",t, s[n]);
else printf("Case %d: %.10f\n", t, log(n) + r + 1.0/(2 * n));
}
return 0;
}
LightOJ - 1214 Large Division
思路
大数除法
代码
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
typedef long long LL;
int main(){
int T;
scanf("%d", &T);
for(int k = 1; k <= T; k ++) {
vector<int> A;
string str;
cin >> str;
LL b;
cin >> b;
if(b < 0) b = -b;
for(int i = 0; i < str.size(); i ++) {
if(str[i] == '-') continue;
else A.push_back(str[i] - '0');
}
LL t = 0;
for(int i = 0; i < A.size(); i ++) {
t = t * 10 + A[i];
t %= b;
}
if(t == 0) printf("Case %d: divisible\n", k);
else printf("Case %d: not divisible\n", k);
}
return 0;
}
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
int cas=1;
while(t-->0)
{
BigInteger a = s.nextBigInteger();
BigInteger b = s.nextBigInteger();
b=b.abs();
if(a.mod(b)==BigInteger.ZERO)
{
System.out.println("Case "+cas+": "+"divisible");
cas++;
}
else
{
System.out.println("Case "+cas+": "+"not divisible");
cas++;
}
}
}
}
LightOJ - 1197Help Hanzo
思路
二次筛法咯。
线性筛 + 埃氏筛法
代码
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1000010;
int primes[N];
int cnt;
bool st[N];
typedef long long LL;
void init(int n) {
memset(st, 0, sizeof st);
cnt = 0;
for(int i = 2; i < N; i ++) {
if(!st[i]) primes[cnt ++] = i;
for(int j = 0; primes[j] * i < N; j ++) {
st[primes[j] * i] = true;
if(i % primes[j] == 0) break;
}
}
}
int main() {
int T;
init(50000);
scanf("%d", &T);
for(int k = 1; k <= T; k ++) {
int l, r;
scanf("%d%d", &l, &r);
memset(st, false, sizeof st);
for(int i = 0; i < cnt; i ++) {
LL p = primes[i];
// 二次筛,筛区间质数, 这里要p*2, 防止结果l等于p，把p也给筛咯
for(LL j = max((l + p - 1) / p * p, p * 2); j <= r; j += p)
//偏移量, 若太大存不住
st[j - l] = true;
}
cnt = 0;
for(int i = 0; i <= r - l; i ++) {
// 把1情况去掉
if(!st[i] && i + l >= 2)
cnt ++;
}
printf("Case %d: %d\n", k, cnt);
}
return 0;
}
LightOJ - 1138 Trailing Zeroes (III)
思路
N！中分解质因子 其中的
5
k
5^k
5k, 质因子5的幂k有多大, 则后面就有多少个0
类似阶乘分解的题。
即
N
!
N!
N!, 计算其中质因子5的幂。
利用 下面代码计算N!中, p的幂s
for(LL i = n; i; i /= p) {
s += i / p;
}
然后利用二分来枚举最优解咯
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL num;
LL solve(LL n) {
LL s = 0;
for(LL i = n; i; i /= 5) {
s += i / 5;
}
return s;
}
int main() {
int T;
scanf("%d", &T);
for(int t = 1; t <= T; t ++) {
scanf("%lld", &num);
LL l = 1, r = 1e12;
while(l < r) {
int mid = l + r >> 1;
if(solve(mid) >= num) r = mid;
else l = mid + 1;
}
if(solve(l) == num)printf("Case %d: %lld\n", t, l);
else printf("Case %d: impossible\n", t);
}
return 0;
}
POJ - 2115 C Looooops
题意
求最小x满足, a x + b ≡ c ( m o d 2 k ) ax + b \equiv c \ (mod \ 2^k) ax+b≡c (mod 2k)
思路
a
+
c
x
=
b
+
y
∗
2
k
a + cx = b + y*2^k
a+cx=b+y∗2k
c
x
−
2
k
∗
y
=
b
−
a
cx - 2^k*y = b - a
cx−2k∗y=b−a
c
x
+
2
k
∗
y
′
=
b
−
a
cx + 2^k*y' = b - a
cx+2k∗y′=b−a
用扩展欧几里得算法求即可
a
x
+
b
y
=
g
c
d
(
a
,
b
)
ax + by = gcd(a,b)
ax+by=gcd(a,b)通解:
{ x = x 0 + b g c d ( a , b ) ∗ t y = y 0 − a g c d ( a , b ) ∗ t \left\{ \begin{aligned} & x = x_0 + \frac{b}{gcd(a, b) } * t \\ & y = y_0 - \frac{a}{gcd(a, b)} * t \end{aligned} \right. ⎩⎪⎪⎨⎪⎪⎧x=x0+gcd(a,b)b∗ty=y0−gcd(a,b)a∗t
代码
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL exgcd(LL a, LL b, LL &x, LL &y) {
if(!b) {
x = 1, y = 0;
return a;
}
LL x1, y1;
LL d = exgcd(b, a % b, x1, y1);
x = y1;
y = x1 - a / b * y1;
return d;
}
int main() {
LL a, b, c, k;
while(~scanf("%lld%lld%lld%lld", &a, &b, &c, &k) && a + b + c + k) {
LL z = (LL)1 << k;
LL x, y;
LL d = exgcd(c, z, x, y);
LL p = b - a;
if(p % d != 0) puts("FOREVER");
else {
x = (LL)x * (p / d);
LL t = z / d;
x = (x % t + t) % t;
printf("%lld\n", x);
}
}
return 0;
}
SGU - 106 The equation
思路
扩展欧几里得+区间交集+细节处理
代码转自：xzx9
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll x1,x2,y1,y2;
ll exgcd(ll a,ll b,ll &x,ll &y)
{
if(!b)
{
x=1;
y=0;
return a;
}
ll ans=exgcd(b,a%b,x,y);
ll tmp=x;
x=y;
y=tmp-(a/b)*y;
return ans;
}
ll cal(ll a,ll b,ll c)
{
ll x,y;
ll gcd=exgcd(a,b,x,y);
if(c%gcd)
return 0;
x=x/gcd*c;
y=y/gcd*c;
int t1 = b / gcd;
int t2 = a / gcd;
//左边界上取整,右边界下取整
ll r1=floor(1.0*(x2-x)/t1);
ll l2=ceil(1.0*(y-y2)/t2);
ll l1=ceil(1.0*(x1-x)/t1);
ll r2=floor(1.0*(y-y1)/t2);
//区间内交集
ll l=max(l1,l2);
ll r=min(r1,r2);
return l<=r?r-l+1:0;
}
int main()
{
ll a,b,c,ans=0;
scanf("%lld%lld%lld",&a,&b,&c);
scanf("%lld%lld",&x1,&x2);
scanf("%lld%lld",&y1,&y2);
if(c>0)//预处理方便处理
c=-c,a=-a,b=-b;
if(a<0)
a=-a,swap(x1,x2),x1=-x1,x2=-x2;
if(b<0)
b=-b,swap(y1,y2),y1=-y1,y2=-y2;
if(x1>x2||y1>y2)//分类讨论
ans=0;
if(a==0&&b==0)
{
if(c==0)
ans=(x2-x1+1)*(y2-y1+1);
}
else if(a==0)
{
ll t=-c/b;
if(b*t+c==0&&t>=y1&&t<=y2)
ans=x2-x1+1;
}
else if(b==0)
{
ll t=-c/a;
if(a*t+c==0&&t>=x1&&t<=x2)
ans=y2-y1+1;
}
else
ans=cal(a,b,-c);
printf("%lld\n",ans);
return 0;
}
UVA - 10200 Prime Time
思路
预处理前缀和咯
注意：
- 精度丢失问题，末尾+1e-8
- L可能为0, 直接取sum[r] 即可
代码
#include<iostream>
using namespace std;
const int N = 1e5 + 10;
int sum[N];
bool is_prime(int x) {
for(int i = 2; i * i <= x; i ++)
if(x % i == 0) return false;
return true;
}
void init() {
for(int i = 0; i <= 10010; i ++) {
if(!i) {sum[i] = 1; continue;}
if(is_prime(i * i + i + 41)) {
sum[i] = sum[i - 1] + 1;
}
else sum[i] = sum[i - 1];
}
}
int main() {
init();
int l, r;
while(~scanf("%d%d", &l, &r)) {
double ans;
if(l == 0) {
ans = 100 * (sum[r]) * 1.0 / (r + 1);
}
else {
ans = 100 * (sum[r] - sum[l - 1]) * 1.0 / (r - l + 1);
}
printf("%.2f\n", (double) ans + 1e-8);
}
return 0;
}
POJ - 2478 Farey Sequence
思路
看一会就知道规律了咯
就是筛欧拉函数模板题咯
代码
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL ;
const int N = 1000010;
int primes[N];
bool st[N];
int cnt;
LL sum[N];
LL phi[N];
typedef long long LL;
void init() {
phi[1] = 1;
for(int i = 2; i < N; i ++) {
if(!st[i]) {
primes[cnt ++] = i;
phi[i] = i - 1;
}
for(int j = 0; (LL)primes[j] * i < N; j ++) {
int t = primes[j] * i;
st[t] = true;
if(i % primes[j] == 0) {
phi[t] = phi[i] * primes[j];
break;
}
phi[t] = phi[i] * (primes[j] - 1);
}
}
for(int i = 1; i < N; i ++)
sum[i] = sum[i - 1] + phi[i];
}
int main() {
int n;
init();
while(~scanf("%d", &n) && n) {
printf("%lld\n", sum[n] - 1);
}
return 0;
}
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