牛客 136J-洋灰三角 +高中数学博大精深

参考学习：http://www.cnblogs.com/l609929321/p/9500814.html

牛客 136J-洋灰三角

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; 
}