参考学习：http://www.cnblogs.com/l609929321/p/9500814.html
题意：
在一个1 * n的棋盘中，第一格放1，之后的每一个放前一个格子的k倍多P个石子，问填满整个棋盘需要多少个石子。
思路：
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <list> #include <cstdlib> #include <iterator> #include <cmath> #include <iomanip> #include <bitset> #include <cctype> using namespace std; //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << " "; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int ,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ' ' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //2147483647 const ll nmos = 0x80000000LL; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18 const double PI=acos(-1.0); template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } // #define _DEBUG; //*// #ifdef _DEBUG freopen("input", "r", stdin); // freopen("output.txt", "w", stdout); #endif /*-----------------------show time----------------------*/ const int mod = 1e9+7; ll ksm (ll a,ll n){ ll res = 1; while(n>0){ if(n&1) res = res * a%mod; a = a * a %mod; n >>= 1; } return res; } int main(){ ll n,k,p; cin>>n>>k>>p; if(k==1){ cout<<(n*(n-1)%mod*ksm(2,mod-2) * p%mod + n)%mod<<endl; return 0; } ll ans = (((ksm(k,n) - 1 + mod)*ksm(k-1,mod-2))%mod*(1 + p * ksm(k-1, mod-2)%mod) + mod)%mod; ans = ((ans - (n*p%mod*ksm(k-1,mod-2))%mod)%mod+mod)%mod; cout<<ans<<endl; return 0; }