首先，今天是汶川地震4周年，深切的悼念！
准备了一个星期的数论，从一开始的野心勃勃，自己去做PPT，自己去选题目，自己去做解题报告。
可是没几天发现自己的精力不够用。我并不需要这些。有的只是去回顾以前学过的内容，挑选好的给大家讲，把自己的经验给大家讲，让大家快点入门。
回想去年我也是在314聆听着谢良的讲座，当时一点也听不懂，到今天能够在这里给别人讲，已是不小的进步。
昨天晚上自己模拟了这场比赛，发现很多知识都不太记得了。编程的熟练度是会下降的，但是数学的思想是不会消褪的，不正是说明还可以在翻腾一遍吗？不是原地踏步，这也是前行。知识的掌握本身就是反复的，而平时的不反复而掌握了只是自欺欺人罢了。
ACM就这样以朴素到极致的方式让你认清你自己，没有任何掩饰。里面浸透的是每一步的付出。在这样一个群体面前，我的讲解只是希望大家能够多交流。ACM的交流很重要啊，这才是ACM比赛的精华之处。通过这次讲解，我更加认识到知识体系的重要性，对大局的把握。一定要总结，就算总结不好，那也是一次自我提炼，这样一个过程才是应该去经历的。
明天就是中南的月赛，好好享受！
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附：
| Problem A | Goldbach's Conjecture | |
| Problem B | X-factor Chains | |
| Problem C | Longge's problem | |
| Problem D | GCD & LCM Inverse | |
| Problem E | 青蛙的约会 | |
| Problem F | Biorhythms | |
| Problem G | Cipher | |
| Problem H | Recurrent Function | |
| Problem I | DP? | |
| Problem J | Least Common Multiple |
| 题目号 |
OJ |
算法 |
题目大意 |
解法描述 |
| 2262 |
poj |
判断素数 |
验证哥德巴赫猜想 |
从小到大枚举较小的一个数，然后判断这个数和减下来的那个数是否都为素数 |
| 3421 |
poj |
质因数分解 |
给出一个数N，从1变到N，最长的长度是多少，有几种方案？ |
最长的长度就是质因子的个数，方案数为排列组合问题，为(t1+t2+…+tn)!/(t1!*t2!*…*tn!) |
| 2480 |
poj |
euler函数 |
|
|
| 2429 |
poj |
GCD&LCM |
给定GCD和LCM的值，求其原值，若有多组答案，则输出和最小的一组 |
需要分解质因数，然后把它搞成 |
| 1061 |
poj |
ExtendedGcd |
两只青蛙按各自的速度跳，问能否在同一时间同一地点相遇？ |
可划归为求ax=b(modn)的形式 |
| 1006 |
poj |
CRT |
|
但是可能的问题是中间的结果会很大， 要当心。 |
| 1026 |
poj |
置换群 |
给出一个置换,以及若干询问，对于每个询问，有k和string，求，k可能很大，的个数不超过200 |
一开始想法是求各个循环的最小公倍数，发现会超大时间也不允许； 由于每个循环独立，故可分别求。 好的方法是先对每个元素算出k次迭代后的f(i)值，然后询问的时候O(1)就可以解决； |
| 3708 |
poj |
置换群+CRT |
递归定义函数 F(j) = a[j] , 1<=j<d F(n*d+j) = d*f(n)+b[j], 0<=j<d, n>=1 {A[I]}={1,2,…,d-1} {b[i]}={0,1,…,d-1} Fx(m)=f(f(f(…f(m)))) x times 求fx(m) = k的最小非负整数x |
F的递归定义是把正整数在d进制下分解 4 = (1 0 0)2 || a[1], b[0], b[0] 所以a只需d-1个元素 X的一次迭代实际上是一个置换群 上式b[0]=1 故F1(4)= (1 1 1)2 = 7 所以可以预处理处一个数组a[i][k] 表示从i到k最少需要几步，同时记录下每个元素的周期cycleA;对b同样处理 可以得到若干个同余方程 X=a1//n1=a2//n2=…=ak//nk,k为d进制下的长度 由于ni不满足两两互质，故需要用CRT的合并方程的方法； 无解的条件： 1。D进制下长度不一样； 2．A[i][k]不存在 3.同余方程无解； |
| 3944 |
hdu |
数论-综合 |
给定n,k,p（0<=k<=n<10^9, P<10000 is a prime) 求在杨辉三角中从(0,0)动(n,k)路径上的最小和 |
公式容易推导为 K = min(k,n-k) C(n+1,k)+(n-k) 接下去就是一个取模的问题了 自己整理出了一个模板 int C(int n, int k, int p); 一个错误检查了一个小时，n!(mod P)在算周期时要用while 去做，因为(1*p),(2*p)…还是有其他的因数的 |
| 1019 |
hdu |
lcm |
求给定N个数的LCM |
|
我自己昨天晚上写的程序，以及以前写的程序，以供参考。
#include <cstdio>
#include <math.h>
using namespace std;
bool check(int n) {
int m = int(sqrt(n));
if (n == 1) return false;
for (int i = 2; i <= m; i++)
if (n % i == 0) return false;
return true;
}
int main() {
int n, i;
while (scanf("%d", &n), n>0) {
i = 1;
while (true) {
if (check(i)&&check(n-i)) break;
i += 2;
}
printf("%d = %d + %d\n", n, i, n-i);
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
//#include <algorithm>
#include <vector>
#include <bitset>
using namespace std;
const int MAXPRIME = 1048576;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<pii>::iterator ITE_pii;
#define mp(x,y) make_pair((x), (y))
#define xx first
#define yy second
bitset<MAXPRIME+10> p;
vector<int> prime;//82100
vector<pii> n_depart;
ll fac[21];
int MakePrime(int L) {
int i, j;
prime.clear(); prime.push_back(2);
p.set();p[0] = p[1] = 0; //make all to 1
for (i = 3; i <= L; i += 2)
if (p[i]) {
prime.push_back(i); //printf("%d\n", i);
if (double(i)>double(L)/i) continue;
for (j = i * i; j <= L; j += i << 1) p[j] = 0;
}
return prime.size();
}
void depart(ll n, vector<pii> &a) {
int cnt;
a.clear();
for (vector<int>::iterator i = prime.begin(); n > 1 && i != prime.end() && (ll)(*i)*(*i) <= n; i++)
if (n % (*i)==0) {
for (cnt = 0; n % (*i) == 0; n /= (*i), cnt++);
a.push_back(mp((*i), cnt));
}
if (n > 1) a.push_back(mp(n, 1));
}
void Get_fac(int n) {
fac[0] = 1;
for (int i = 1; i <= n; i++) fac[i] = fac[i-1]*i;
}
ll n, ans;
int tot;
int main() {
MakePrime(MAXPRIME);
Get_fac(20);
while (scanf("%I64d", &n) == 1) {
depart(n, n_depart);
ans = 1; tot = 0;
for (ITE_pii k = n_depart.begin(); k != n_depart.end(); k++) {
tot += k->yy;
ans *= fac[k->yy];
}
ans = fac[tot]/ans;
printf("%d %I64d\n", tot, ans);
}
return 0;
}
C: Longge's problem
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>
#include <limits>
using namespace std;
typedef long long ll;
int main() {
setbuf(stdout, NULL);
ll N;
ll p;
ll a;
ll res;
while (cin >> N) {
res = N;
for (ll i = 2; i * i <= N; i++) {//这里求N得标准质因数分解很好，学会了。。
if (N % i == 0) {
p = i;
a = 0;
while (N % p == 0) {
a++;
N /= p;
}
res = res + res * a * (p - 1) / p;
}
}
if (N != 1) {
res = res * (2 * N - 1) / N;
}
cout << res << endl;
}
return 0;
}
D: GCD & LCM Inverse
#include<ctime>
#include<iostream>
#include<fstream>
#include <cstdio>
#include <algorithm>
using namespace std;
long long factor[1000],fac_top = -1;
//计算两个数的gcd
long long gcd(long long a,long long b)
{
if(a==0)
return b;
long long c;
while(b!=0)
{
c=b;
b=a%b;
a=c;
}
return a;
}
//ret = (a*b)%n (n<2^62)
long long muti_mod(long long a,long long b,long long n)
{
long long exp = a%n, res = 0;
while(b)
{
if(b&1)
{
res += exp;
if(res>n) res -= n;
}
exp <<= 1;
if(exp>n)
exp -= n;
b>>=1;
}
return res;
}
// ret = (a^b)%n
long long mod_exp(long long a,long long p,long long m)
{
long long exp=a%m, res=1; //
while(p>1)
{
if(p&1)//
res=muti_mod(res,exp,m);
exp = muti_mod(exp,exp,m);
p>>=1;
}
return muti_mod(res,exp,m);
}
//miller-rabin法测试素数, time 测试次数
bool miller_rabin(long long n, long long times)
{
if(n==2)return 1;
if(n<2||!(n&1))return 0;
long long a, u=n-1, x, y;
int t=0;
while(u%2==0){
t++;
u/=2;
}
srand(time(0));
for(int i=0;i<times;i++)
{
a = rand() % (n-1) + 1;
x = mod_exp(a, u, n);
for(int j=0;j<t;j++)
{
y = muti_mod(x, x, n);
if ( y == 1 && x != 1 && x != n-1 )
return false; //must not
x = y;
}
if( y!=1) return false;
}
return true;
}
long long pollard_rho(long long n,int c)
{//找出一个因子
long long x,y,d,i = 1,k = 2;
srand(time(0));
x = rand()%(n-1)+1;
y = x;
while(true) {
i++;
x = (muti_mod(x,x,n) + c) % n;
d = gcd(y-x, n);
if(1 < d && d < n)return d;
if( y == x) return n;
if(i == k) {
y = x;
k <<= 1;
}
}
}
void findFactor(long long n,int k)
{//二分找出所有质因子，存入factor
if(n==1)return;
if(miller_rabin(n, 10))
{
factor[++fac_top] = n;
return;
}
long long p = n;
while(p >= n)
p = pollard_rho(p,k--);//k值变化，防止死循环
findFactor(p,k);
findFactor(n/p,k);
}
long long a[1000];
long long m,minx,ans;
int cmp(const void *a,const void *b){
return *(long long *)a-*(long long *)b;
}
void dfs(long long s,long long num,long long t){
if(s==m+1)
{
if(minx==-1||(num+t/num<minx))
{
minx=num+t/num;
ans=num;
}
return;
}
dfs(s+1,num*a[s],t);
dfs(s+1,num,t);
}
void solve(){
qsort(factor,fac_top+1,sizeof(long long),cmp);
a[0]=factor[0];
long long i;
for(i=0;i<fac_top;i++)
if(factor[i]==factor[i+1])
{
a[m]*=factor[i+1];
}
else
{
m++;
a[m]=factor[i+1];
}
}
int main()
{
long long s,t,n;
// ifstream cin("in.txt");
while(cin>>s>>t)
{
n=t/s;
if(s==t)
{
cout<<s<<' '<<t<<endl;
continue;
}
minx=-1;
m=0;
fac_top = -1;
findFactor(n,107);
solve();
dfs(0,1,n);
if(ans>n/ans) ans=n/ans;
cout<<s*ans<<' '<<s*(n/ans)<<endl;
}
return 0;
}
//http://www.cnblogs.com/zhaozhe/archive/2011/04/12/2013979.html
E: 青蛙的约会
#include <stdio.h>
long long exgcd(long long a, long long b, long long &x, long long &y) {
if (b == 0) {
x = 1; y = 0; return a;
}
long long g = exgcd(b, a % b, x, y);
long long t = x - (a / b) * y;
x = y;
y = t;
return g;
}
int main() {
long long x, y, m, n, l, x0, y0, a, b, c, d, t;
scanf("%I64d%I64d%I64d%I64d%I64d", &x, &y, &m, &n, &l);
a = n-m; b = l; c = x-y;
if (a < 0) { a = -a; c = -c; }
long long g = exgcd(a, b, x0, y0);
if (n == m || (x - y) % g != 0) printf("Impossible\n"); else {
b /= g;
c /= g;
t = c * x0;
printf("%I64d\n", (t % b+b) % b);
}
return 0;
}
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 3;
LL n[maxn] = {23,28,33};
LL a[maxn], m[maxn], mi[maxn];
LL N, pi;
//ax+by=gcd(a,b)
LL exgcd(LL a, LL b, LL &x, LL &y) {
if (b == 0) {
x = 1; y = 0;
return a;
};
LL g = exgcd(b, a%b, x, y);
LL t = x-a/b*y;
x = y;
y = t;
return g;
}
//ax=1(mod n), if gcd(a,n)!=1 inverse do not exist
LL inverse(LL a, LL n) {
LL d, x, y;
d = exgcd(a, n, x, y);
if (d != 1) return -1;
x = (x%n+n)%n;
return x;
}
LL cases, d, ans;
int main() {
N = 3;
cases = 0;
while (true) {
pi = 1;
for (int i = 0; i < N; i++) {
scanf("%lld", &a[i]);
if (a[i]==-1) return 0;
pi = pi*n[i];
}
ans = 0;
for (int i = 0; i < N; i++) {
m[i] = pi/n[i];
mi[i] = inverse(m[i], n[i]);
ans = (ans+a[i]*m[i]*mi[i]%pi)%pi;
}
scanf("%lld", &d);
ans = (ans-d+pi-1)%pi+1;
printf("Case %lld: the next triple peak occurs in %lld days.\n", ++cases, ans);
}
}
G:Cipher
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
const int maxn = 210;
int n, next[maxn], k;
int d[maxn][maxn], ds[maxn];
char str[maxn], ans[maxn];
int i, j;
int main() {
while (scanf("%d", &n), n>0) {
for (i = 0; i < n; i++) scanf("%d", &next[i]);
for (i = 0; i < n; i++) next[i]--;
for (i = 0; i < n; i++) {
d[i][0] = i; ds[i] = 1;
for (j = next[i]; j != i; j = next[j])
d[i][ds[i]++] = j;
}
while (scanf("%d", &k), k>0) {
getchar();
gets(str);
i = 0; while (str[i]) i++; while (i<n) str[i++]=' ';
for (i = 0; i < n; i++) ans[d[i][k%ds[i]]] = str[i];
ans[n] = 0;
puts(ans);
}
printf("\n");
}
return 0;
}
/*
* Author: Zhang Jiangbin
* Created Time: 2011/8/20/星期六 12:52:34
* File Name: a.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
#define SZ(v) ((int)(v).size())
typedef __int64 LL;
const int maxd = 105;
const int maxn = 405;
LL MODa[maxn], MODn[maxn];
LL MODN;
LL exgcd(LL a, LL b, LL &x, LL &y);
LL inverse(LL a, LL n);
LL ChineseRemainderTheorem(LL N, LL a[], LL n[]);
void prepare();
void read(int c[], int&);
int divide(int &L, int c[], int d);
int a[maxd], b[maxd];
int d, i, j;
int A[maxd][maxd], B[maxd][maxd], cycleA[maxd], cycleB[maxd];
char st[maxd];
int nu[maxd];
int c1[maxn], c2[maxn], len1, len2;//lg2(10^100)<400
bool ok;
int t;
LL ans;
int main()
{
//freopen("a.in", "r", stdin);
while (scanf("%d", &d), d>=0) {
for (i = 1; i < d; i++)
scanf("%d", a+i);
for (i = 0; i < d; i++)
scanf("%d", b+i);
prepare();
read(c1, len1);
read(c2, len2);
if (len1!=len2)
puts("NO");
else {
ans = 1;
ok = true;
MODN = 0;
for (i = 0; i < len1-1; i++){
t = B[c1[i]][c2[i]];
if (t==-1) {
ok = false;
break;
}
MODa[MODN] = t; MODn[MODN] = cycleB[c1[i]];
MODN++;
}
t = A[c1[i]][c2[i]];
if (t == -1){
ok = false;
} else {
MODa[MODN] = t; MODn[MODN] = cycleA[c1[i]];
MODN++;
}
if (ok) {
ans = ChineseRemainderTheorem(MODN, MODa, MODn);
if (ans==-1) ok = false;
}
if (!ok) puts("NO"); else printf("%I64d\n", ans);
}
}
return 0;
}
void prepare(){
int i, j, t;
memset(A,-1,sizeof(A));
memset(B,-1,sizeof(B));
for (i = 1; i < d; i++){
A[i][i] = 0;
t = i; //printf("t:%d ", t);
for (j = 1; j <= d; j++){
t = a[t];//printf("t:%d ", t);
if (t == i) {
cycleA[i] = j;
break;
}
A[i][t] = j;
}
// printf("A : %d : cycle : %d\n", i, cycleA[i]);
}
for (i = 0; i < d; i++){
B[i][i] = 0;
t = i;//printf("t:%d ", t);
for (j = 1; j <= d; j++){
t = b[t];//printf("t:%d ", t);
if (t == i){
cycleB[i] = j;
break;
}
B[i][t] = j;
}
// printf("B : %d : cycle : %d\n", i, cycleB[i]);
}
//debug
}
void read(int c[], int &len) {
int l, num, i, j;
scanf("%s", st);
// printf("%s\n", st);
l = strlen(st);
memset(nu,0,sizeof(nu));
for (i = 0, j = l-1; i < l; i++, j--)
nu[i] = st[j]-'0';
len = 0;
do{
c[len++] = divide(l, nu, d);
// printf("%d ", c[len-1]);
} while(l);
// printf("\n");
}
int divide(int &l, int c[], int d) {
int num, i;
num = 0;
for (i = l-1; i >= 0; i--){
num = num*10+c[i];
c[i] = num/d;
num %= d;
}
while (l&&(!c[l-1])) l--;
return num;
}
//ax+by=gcd(a,b)
LL exgcd(LL a, LL b, LL &x, LL &y) {
if (b == 0) {
x = 1; y = 0;
return a;
};
LL g = exgcd(b, a%b, x, y);
LL t = x-a/b*y;
x = y;
y = t;
return g;
}
//ax=1(mod n), if gcd(a,n)!=1 inverse do not exist
LL inverse(LL a, LL n) {
LL d, x, y;
d = exgcd(a, n, x, y);
if (d != 1) return -1;
x = (x%n+n)%n;
return x;
}
//方程的个数N， 形式为 x=a(i)(mod n(i))
//无解返回-1
LL ChineseRemainderTheorem(LL N, LL a[], LL n[]) {
LL d, x, y, c;
for (LL i = 0; i < N; i++)
if (a[i] >= n[i]) return -1;
for (LL i = 1; i < N; i++) {
//n[i-1]*x=(a[i]-a[i-1]) (mod n[i])
d = exgcd(n[i-1], n[i], x, y);
if ((a[i]-a[i-1])%d!=0) return -1;
x = x*((a[i]-a[i-1])/d);
x = (x%n[i]+n[i])%n[i];//non-negative
x = x%(n[i]/d);//should get the minest x, actually it needn't for it doesn't matter a[i]
//calculate new a and n
c = a[i-1]+x*n[i-1];
//c = a[i-1]+(x+i*n[i]/d)*n[i-1]=a[i-1]+x*n[i-1]+i*(n'[i]);
n[i] = n[i]/d*n[i-1];
a[i] = c%n[i];
}
return a[N-1];
}
/*
* Author: Zhang Jiangbin
* Created Time: 2011/8/16/星期二 12:48:42
* File Name: e.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define SZ(v) ((int)(v).size())
const int maxp = 10005;
#define maxprime 10001//9973
#define primes 1240//1229
int f[primes][maxp], inv[primes][maxp];
int n, k ,p;
int makeprime(int L);
void prepare();
int exgcd(int a, int b, int &x, int &y);
int inverse(int a, int n);//求a对于质数n的逆元
int get(int n, int p);//求n!中质因子p的个数
int quick(int a, int b, int p);//乘法快速幂
bool ph[maxprime+10];//1 is a prime;0 is not
int prime[primes], primelen;//prime[i] is the i-th prime, 0..primelen-1
int getf(int pos, int n) {
int t1 = quick(f[pos][p-1],n/p, p);
if (n%p) t1 = (t1*f[pos][n%p])%p;
return t1;
}
int getinv(int pos, int n) {
int t1 = quick(inv[pos][p-1],n/p, p);
if (n%p) t1 = (t1*inv[pos][n%p])%p;
return t1;
}
int C(int n, int k, int p) {
int numofp = get(n, p)-get(k, p)-get(n-k, p);
if (numofp>0) return 0;
//binary search
int l = 0, r = primelen;
while (l+1 < r) {
int mid = (l+r)/2;
if (p>=prime[mid]) l = mid; else r = mid;
}
int pos = l;
// printf("%d %d\n", prime[pos], p);
int ans = 1, nn;
nn = n;
while (nn>0) {
ans = (ans*getf(pos,nn))%p;
nn/=p;
}
nn = k;
while (nn>0) {
ans = (ans*getinv(pos,nn))%p;
nn/=p;
}
nn = n-k;
while (nn>0) {
ans = (ans*getinv(pos, nn))%p;
nn/=p;
}
return ans;
}
int main()
{
prepare();
int cases = 0;
while (scanf("%d%d%d", &n, &k, &p) == 3) {
k = max(k, n-k);
int ans = C(n+1, k+1, p);
// printf("%d\n", ans);
ans += k;
ans = (ans%p+p)%p;
printf("Case #%d: %d\n", ++cases, ans);
}
return 0;
}
/*求逆元*/
//ax+by=gcd(a,b)
int exgcd(int a, int b, int &x, int &y) {
if (b == 0) {
x = 1; y = 0;
return a;
};
int g = exgcd(b, a%b, x, y);
int t = x-a/b*y;
x = y;
y = t;
return g;
}
//ax=1(mod n), if gcd(a,n)!=1 inverse do not exist
int inverse(int a, int n) {
int d, x, y;
d = exgcd(a, n, x, y);
if (d != 1) return -1;
x = (x%n+n)%n;
return x;
}
//构造小于等于L的素数表
//ph[i]=1为素数 ph[i]=0不是素数，注意1
//prime[i],0..primelen-1为素数表
int makeprime(int L) {
memset(ph,1,sizeof(ph));
primelen = 0;
ph[1] = 0;
for (int i = 2; i <= L; i++)
if (ph[i]) {
prime[primelen++] = i;
//printf("%d\n", i);
for (int j = i+i; j <= L; j += i) ph[j] = 0;
}
return primelen;
}
void prepare() {
makeprime(10000);
// printf("%d\n", prime[primelen-1]);
for (int k = 0; k < primelen; k++) {
int p = prime[k];
f[k][0] = f[k][1] = 1;
inv[k][0] = inv[k][1] = 1;
for (int i = 2; i < p; i++) {
f[k][i] = (f[k][i-1]*i)%p;
inv[k][i] = (inv[k][i-1]*inverse(i,p))%p;
}
// printf("%d %d\n", p, f[k][p-1]);
}
}
//cnt = n! 个p
int get(int n, int p) {
int cnt = 0;
while (n>0) {
cnt += n/p;
n/=p;
}
return cnt;
}
int quick(int a, int b, int p) {
int ans = 1, tmp = a%p;
while (b>0) {
if (b & 1) ans = (ans*tmp)%p;
b/=2;
tmp = (tmp*tmp)%p;
}
return ans;
}
#include <iostream>
using namespace std;
int gcd(int a, int b) {
if (b == 0) return a; else return gcd(b, a % b);
}
int LCM(int a, int b) {
int c = gcd(a, b);
return a / c * b;
}
int c, n, lcm;
int a[1000];
int main(){
scanf("%d", &c);
while (c--) {
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int lcm = a[1];
for (int i = 2; i <= n; i++)
lcm = LCM(a[i], lcm);
printf("%d\n", lcm);
}
return 0;
}