題目鏈接：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 S_i?= Aⁱ?mod B and T_i?= Min{ S_k?| i-A <= k <= i, k >= 1}
Your task is to calculate the product of T_i?(1 <= i <= n) mod B.

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

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

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Sample Input 1 2 3 2 3 4 3 4 5 4 5 6 5 6 7

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Sample Output 2 3 4 5 6

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Author WhereIsHeroFrom@HDU

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Source HDOJ 5th Anniversary Contest 單調(diào)隊列簡單題； 卡內(nèi)存，list過不了，deque，G++過的； #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; /// 數(shù)組大小 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; }

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轉(zhuǎn)載于:https://www.cnblogs.com/jhz033/p/6666204.html

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