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 /**************************************************************************************/
