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题目链接:
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
/************************************************ ┆ ┏┓ ┏┓ ┆ ┆┏┛┻━━━┛┻┓ ┆ ┆┃ ┃ ┆ ┆┃ ━ ┃ ┆ ┆┃ ┳┛ ┗┳ ┃ ┆ ┆┃ ┃ ┆ ┆┃ ┻ ┃ ┆ ┆┗━┓ ┏━┛ ┆ ┆ ┃ ┃ ┆ ┆ ┃ ┗━━━┓ ┆ ┆ ┃ AC代马 ┣┓┆ ┆ ┃ ┏┛┆ ┆ ┗┓┓┏━┳┓┏┛ ┆ ┆ ┃┫┫ ┃┫┫ ┆ ┆ ┗┻┛ ┗┻┛ ┆ ************************************************ */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e6+10; const int maxn=2e3+14; const double eps=1e-12; LL pow_mod(LL x,LL y) { LL s=1,base=x; while(y) { if(y&1)s=s*base%mod; base=base*base%mod; y>>=1; } return s; } int main() { int t; read(t); while(t--) { LL n,m; read(n);read(m); cout<<(pow_mod(m,n+1)-1+mod)%mod*pow_mod(m-1,mod-2)%mod<<"\n"; } return 0; }
hdu-5793 A Boring Question(二项式定理)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5737926.html