标签：style http color os io ar for cti sp
Young cryptoanalyst Georgie is investigating different schemes of generating random integer numbers ranging from 0 to m - 1. He thinks that standard random number generators are not good enough, so he has invented his own scheme that is intended to bring more randomness into the generated numbers. First, Georgie chooses n and generates n random integer numbers ranging from 0 to m - 1. Let the numbers generated be a1, a2,..., an. After that Georgie calculates the sums of all pairs of adjacent numbers, and replaces the initial array with the array of sums, thus getting n - 1 numbers: a1 + a2, a2 + a3,..., an - 1 + an. Then he applies the same procedure to the new array, getting n - 2 numbers. The procedure is repeated until only one number is left. This number is then taken modulo m. That gives the result of the generating procedure. Georgie has proudly presented this scheme to his computer science teacher, but was pointed out that the scheme has many drawbacks. One important drawback is the fact that the result of the procedure sometimes does not even depend on some of the initially generated numbers. For example, if n = 3 and m = 2, then the result does not depend on a2. Now Georgie wants to investigate this phenomenon. He calls the i-th element of the initial array irrelevant if the result of the generating procedure does not depend on ai. He considers various n and m and wonders which elements are irrelevant for these parameters. Help him to find it out.
n100 000, 2m109)
in a single line.
3 2
1 2
题意：整个式子的和可以 化简为 sigma (C(n-1,i-1)*ai)
思路：只要判断C(n-1,i-1)能否被 m整除即可。
做法是先分解m的质因数，然后计算1！~(n-1)! 包含m的质因数的个数
C（n-1,i-1） = (n-1)!/((i-1)!*(n-i)!)
只要判断 剩下的质因数的个数是否大于等于m的任一个质因数的个数即可
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
typedef long long LL;
const int maxp = 40000+10;
const int maxn = 100000+10;
int n,m;
bool isPrime[maxp];
vector<int> ret,prime,hp,cnt,tmp;
vector<int> g[maxn];
void getPrime(){
memset(isPrime,true,sizeof isPrime);
for(int i = 2; i < maxp; i++){
if(isPrime[i]){
prime.push_back(i);
for(int j = i+i;j < maxp; j += i){
isPrime[j] = false;
}
}
}
}
void getDigit(){
int tk = m;
for(int i = 0; i < prime.size() && prime[i]*prime[i] <= tk; i++){
if(tk%prime[i]==0){
int k = 0;
hp.push_back(prime[i]);
while(tk%prime[i]==0){
tk /= prime[i];
k++;
}
cnt.push_back(k);
}
}
if(tk>1){
hp.push_back(tk);
cnt.push_back(1);
}
}
void init(){
cnt.clear();
hp.clear();
ret.clear();
getDigit();
for(int i = 0; i <= n-1; i++) g[i].clear();
for(int i = 0; i <= n-1; i++) {
for(int j = 0; j < hp.size(); j++){
int d = 0,t = i;
while(t) {
d += t/hp[j];
t /= hp[j];
}
g[i].push_back(d);
}
}
}
void solve(){
bool miden = false;
for(int i = 2; i <= (n-1)/2+1; i++){
bool flag = true;
for(int j = 0; j < hp.size(); j++){
int d = g[n-1][j]-g[i-1][j]-g[n-i][j];
if(d < cnt[j]){
flag = false;
break;
}
}
if(flag) {
ret.push_back(i);
if(i==(n-1)/2+1 && (n&1)) miden = true;
}
}
tmp.clear();
tmp = ret;
if(n&1){
int i;
if(miden) i = tmp.size()-2;
else i = tmp.size()-1;
for(; i >= 0; i--){
ret.push_back(n+1-tmp[i]);
}
}else{
for(int i = tmp.size()-1; i >= 0; i--){
ret.push_back(n+1-tmp[i]);
}
}
printf("%d\n",ret.size());
if(ret.size()){
printf("%d",ret[0]);
for(int i = 1; i < ret.size(); i++){
printf(" %d",ret[i]);
}
}
puts("");
}
int main(){
//freopen("test.txt","r",stdin);
getPrime();
while(~scanf("%d%d",&n,&m)){
init();
solve();
}
return 0;
}UVa1635 - Irrelevant Elements（质因数分解）
标签：style http color os io ar for cti sp
原文地址：http://blog.csdn.net/mowayao/article/details/38930127
