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hdu 3706 Second My Problem First 单调队列

2024-09-12 11:34:52   213人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3706

Second My Problem First

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)



Problem Description
Give you three integers n, A and B. 
Then we define Si = Ai mod B and Ti = Min{ Sk | i-A <= k <= i, k >= 1}
Your task is to calculate the product of Ti (1 <= i <= n) mod B.
 

 

Input
Each line will contain three integers n(1 <= n <= 107),A and B(1 <= A, B <= 231-1). 
Process to end of file.
 

 

Output
For each case, output the answer in a single line.
 

 

Sample Input
1 2 32 3 43 4 54 5 65 6 7
 

 

Sample Output
23456
 

 

Author
WhereIsHeroFrom@HDU
 

 

Source
HDOJ 5th Anniversary Contest
 
单调队列简单题；
卡内存，list过不了，deque，G++过的；
#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cmath>#include<string>#include<queue>#include<algorithm>#include<stack>#include<cstring>#include<vector>#include<list>#include<set>#include<map>using namespace std;#define ll long long#define pi (4*atan(1.0))#define eps 1e-14#define bug(x)  cout<<"bug"<<x<<endl;const int N=1e7+10,M=4e6+10,inf=2147483647;const ll INF=1e18+10,mod=1e9+7;///   数组大小int d[N];ll num[N];int main(){    ll n,a,b;    while(~scanf("%lld%lld%lld",&n,&a,&b))    {        ll ans=1,base=a%b;        int s=0,e=0;        for(int i=1;i<=n;i++,base=(base*a)%b)        {            num[i]=base;            while(s<e&&d[s]<i-a)s++;            while(e>s&&num[d[e]]>=num[i])e--;            d[++e]=i;            //cout<<num[d[s+1]]<<" "<<endl;            ans=ans*(num[d[s+1]])%b;        }        printf("%lld\n",ans);    }    return 0;}

 

 

hdu 3706 Second My Problem First 单调队列


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