Second My Problem First HDU
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题目链接
https://vjudge.net/problem/HDU-3706
题目大意
给你n, A, B;
让你算S i = A i mod B, T i = Min{ S k | i-A <= k <= i, k >= 1}
T i (1 <= i <= n) mod B
核心思想
裸的单调队列题
单调队列可以在o(n)的时间复杂度计算出区间最大值或最小值
(所以如果不用区间求和之类的,单调队列会比线段树少一个log)
具体关于单调队列的知识这个博客讲的比较详细:http://blog.csdn.net/justmeh/article/details/5844650
代码
数组模拟
#include<bits/stdc++.h>using namespace std;typedef pair<int, int> P;typedef long long ll;P cnt[100005];//first存数值, second存idint main() { ll n,a,b; while(scanf("%lld %lld %lld",&n,&a,&b)!=EOF) { int k1=1,k2=1; //队列头尾 long long ans=1,s=1; for(int i=1;i<=n;i++) { s=(s*a)%b; while(k1!=k2 && cnt[k2-1].first>s) --k2;//维护单调队列 cnt[k2].second=i,cnt[k2++].first=s; //入队 while(i-cnt[k1].second>a) ++k1; //元素个数大于a个,出队 ans*=cnt[k1].first; //乘积 ans%=b; } printf("%lld\n",ans); } return 0; }
用stl 的双端队列
#include<bits/stdc++.h>using namespace std;typedef pair<int, int> P;const int INF = 1e9 + 7;deque<P>cnt;int main(){ long long n, a, b; while(~scanf("%lld%lld%lld", &n, &a, &b)) { cnt.clear(); long long ans = 1, s = 1; for(int k = 1; k <= n; ++k) { s = (s*a)%b; while(!cnt.empty() && cnt.back().first > s) cnt.pop_back(); cnt.push_back(P(s, k)); if(cnt.back().second < k - a) cnt.pop_front(); ans*=cnt.front().first; ans%=b; } cout<<ans<<endl; } return 0;}
心得以及注意点
- 两种方法时间差不多,说明deque效率还是可以的,但是如果这题id也用deque来维护,那就要超时了
- 注意爆int啊,一定要注意开long long啊
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