upc 9318 Slot Machines-白红宇

upc 9318 Slot Machines

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发布时间：2019-06-13

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Slot Machines

时间限制: 1 Sec 内存限制: 128 MB

提交: 198 解决: 53

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题目描述

Slot machines are popular game machines in casinos. The slot machine we are considering has six places where a figure appears. By combination of figures, one may earn or lose money. There are ten kinds of figures, so we will represent a figure with a number between 0 and 9. Then we can use a six-digit number w = w1w2w3w4w5w6 where 0 ≤ w1, w2, w3, w4, w5, w6 ≤ 9 to represent one possible outcome of the slot machine.

Figure I.1. The layout of a slot machine.

Old slot machines were made up with mechanical components, but nowadays they were replaced by PC-based systems. This change made one critical flaw: they are based on pseudo-random number generators and the outcome sequences of a slot machine are periodic. Let T[i] be the i-th outcome of a slot machine. At first, there is a truly random sequence of length k,T[1],T[2],…,T[k]. Then there exists one positive number such that T[i+p] = T[i]for all possible values of i(>k). Once an attacker can find out the exact values of k and p, he or she can exploit this fact to beat the casino by betting a lot of money when he or she knows the outcome with a good combination in advance.

For example, you have first six numbers of outcome sequences: 612534, 3157, 423, 3157, 423, and 3157. Note that we can remove first 0’s. Therefore, 3157 represents 003157 and 423 represents 000423. You want to know its tenth number. If you know the exact values of k and p, then you can predict the tenth number. However, there are many candidates for k and p: one extreme case is k=5 and p=1, and another is k=0 and p=6. The most probable candidate is the one where both k and p are small. So, our choice is the one with the smallest k+p. If there are two or more such pairs, we pick the one where p is the smallest. With our example, after some tedious computation, we get k=1 and p=2.

Assume that you have n consecutive outcomes of a slot machine, T[1], T[2], …, T[n]. Write a program to compute the values of k and p satisfying the above-mentioned condition.

输入

Your program is to read from standard input. The first line contains a positive integer n (1 ≤ n ≤ 1,000,000), representing the length of numbers we have observed up to now in the outcome sequence. The following line contains n numbers. Each of these numbers is between zero and 999,999.

输出

Your program is to write to standard output. Print two integers k and p in one line.

样例输入

6612534 3157 423 3157 423 3157

样例输出

1 2

题意

给n个数，删去p个数后会形成循环，最小循环节为k，求p+k最小的解。

分析

用KMP直接求最小循环节即可。

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                    #define INF 0x3f3f3f3f#define FAST_IO ios::sync_with_stdio(false)const double PI = acos(-1.0);const double eps = 1e-6;const int MAX=1e6+10;const int mod=1e9+7;typedef long long ll;using namespace std;#define gcd(a,b) __gcd(a,b)inline ll lcm(ll a,ll b){ return a/gcd(a,b)*b;}inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){ if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}inline ll inv1(ll b){ return qpow(b,mod-2);}inline ll exgcd(ll a,ll b,ll &x,ll &y){ if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;}inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-'0';return x*f;}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );ll s[MAX],n;ll Next[MAX];void KMP(){ ll i,j; i=1,j=0; Next[1]=0; while(i<=n) { if(j==0 || s[i]==s[j]) i++,j++,Next[i]=j; else j=Next[j]; } return ;}int main(){ scanf("%lld",&n); for(ll i=n;i>=1;i--) scanf("%lld",&s[i]); KMP(); ll tp,tk,k,p; k=INF;p=INF; for(int i=1;i<=n;i++) { tp=i-Next[i+1]+1; tk=n-i; if(tp+tk

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转载于:https://www.cnblogs.com/Kissheart/p/9744373.html

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