标签:ott proc math map 单调队列 tom ack list arc
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3706
Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
#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
标签:ott proc math map 单调队列 tom ack list arc
原文地址:http://www.cnblogs.com/jhz033/p/6666204.html