微积分题--HDU6472--艺术台阶

题意：http://acm.hdu.edu.cn/showproblem.php?pid=6472

rt。

思路：https://www.cnblogs.com/graytido/p/12296710.html

差不多了

  1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include <cstdio>//sprintf islower isupper
  3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
  4 #include <iostream>//pair
  5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin);
  6 #include <bitset>
  7 //#include <map>
  8 //#include<unordered_map>   http://acm.hdu.edu.cn/showproblem.php?pid=6472
  9 #include <vector>
 10 #include <stack>
 11 #include <set>
 12 #include <string.h>//strstr substr strcat
 13 #include <string>
 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 18 #include <vector>//emplace_back
 19 //#include <math.h>
 20 #include <cassert>
 21 #include <iomanip>
 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 25 //******************
 26 clock_t __START,__END;
 27 double __TOTALTIME;
 28 void _MS(){__START=clock();}
 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 30 //***********************
 31 #define rint register int
 32 #define fo(a,b,c) for(rint a=b;a<=c;++a)
 33 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 34 #define mem(a,b) memset(a,b,sizeof(a))
 35 #define pr printf
 36 #define sc scanf
 37 #define ls rt<<1
 38 #define rs rt<<1|1
 39 typedef pair<int,int> PII;
 40 typedef vector<int> VI;
 41 typedef unsigned long long ull;
 42 typedef long long ll;
 43 typedef double db;
 44 const db E=2.718281828;
 45 const db PI=acos(-1.0);
 46 const ll INF=(1LL<<60);
 47 const int inf=(1<<30);
 48 const db ESP=1e-9;
 49 const int mod=(int)1e9+7;
 50 const int N=(int)1e6+10;
 51 ll qpow(ll a,ll b,ll mod){
 52     ll ans;
 53 //    a%=mod;
 54     ans=1;
 55     while(b!=0)
 56     {
 57         if(b&1)
 58             ans=(ans*a)%mod;
 59         b/=2;
 60         a=(a*a)%mod;
 61     }
 62     return ans;
 63 }
 64 
 65 ll Div[N];
 66 ll a[N];
 67 struct node
 68 {
 69     ll xi,zhi;
 70 };
 71 vector<node>v;
 72 
 73 void solve()
 74 {
 75     v.clear();
 76     int n;
 77     ll down=1;
 78     sc("%d",&n);
 79     for(int i=1;i<=n;++i)sc("%lld",&a[i]),down=down*Div[a[i]]%mod;
 80     for(int i=n-1;i>=1;--i)a[i]=min(a[i],a[i+1]);
 81     v.push_back({1,0});//系数1,指数0
 82     for(int i=n;i>=1;--i)
 83     {
 84         int sz=v.size();
 85         ll ans=0;
 86         for(int j=0;j<sz;++j)
 87         {
 88             v[j].zhi++;
 89             v[j].xi=(v[j].xi*Div[v[j].zhi])%mod;
 90             ans=(ans+(qpow(a[i],v[j].zhi,mod)*v[j].xi%mod+mod)%mod)%mod;
 91             v[j].xi=-v[j].xi;//变号
 92         }
 93         v.push_back({ans,0});
 94     }
 95     int sz=v.size();//400000003
 96     pr("%lld\n",((v[sz-1].xi*down)%mod+mod)%mod);
 97 }
 98 
 99 int main()
100 {
101     for(int i=1;i<=2005;++i)Div[i]=qpow(i,mod-2,mod);
102     pr("%lld\n",1ll*25*Div[2]%mod);
103     pr("%lld\n",1ll*25*Div[2]%mod*Div[25]%mod);
104     int T;
105     sc("%d",&T);
106     while(T--)solve();
107     return 0;
108 }
109 
110 /**************************************************************************************/