梦回吹角连营(2)(快速幂快乘)

Description

给定f(n)=(a+1)*n^a+(a+2)*n^(a+1)+...+b*n^(b-1)
求f(n)%10000000033

Input

输入一个正整数T(T<=10),表示有T组数据，每组数据包括三个整数a,b,n
(0<=n<=10^9,1<=a <= b-1<=10^20)

Output

输出 f(n)%10000000033 的结果

Sample Input

1
1 2 3

Sample Output

思路：

化简一下式子，得到：

得到 S(a),S(b)，用快速幂模板

注意：这题数非常大，两数相乘用快乘模板。

代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<math.h>
#include<map>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
#define per(i,a,b) for(int i=a;i<=b;i++)
#define ber(i,a,b) for(int i=a;i>=b;i--)
const int N = 1e5 + 5;
const LL mod = 10000000033;
long long n, a, b,aa,bb;
char sa[33], sb[33];
LL mul(LL a, LL b, LL mod)
{
   LL ans = 0;
   while (b)
   {
       if (b & 1)
           ans =(ans+a)%mod;
       b >>= 1;
       a = (a + a) % mod;
   }
   return ans;
}
LL quick(LL a, LL b, LL mod)
{
   LL ans = 1;
   while (b)
   {
       if (b & 1)
       ans = mul(ans, a,mod);
       b >>= 1;
       a = mul(a, a,mod);
   }
   return ans;
}
void into()
{
   int lena = (int)strlen(sa+1), lenb = (int)strlen(sb + 1);
   per(i, 1, lena)
   {
       a = (a*10 + sa[i] - '0') % mod;
       aa = (aa*10 + sa[i] - '0') % (mod - 1);
   }
   per(i, 1, lenb)
   {
       b = (b*10 + sb[i] - '0') % mod;
       bb = (bb*10 +sb[i] -'0') % (mod - 1);
   }
}
int main()
{
   int T;
   cin >> T;
   while (T--)
   {
       a = aa = b = bb = 0;
       cin >> sa+1 >> sb+1 >> n;
       into();
       if (n == 0)
       {
           cout << 0 << endl;
           continue;
       }
       if (n == 1)
       {
           LL ni = quick(2, mod - 2, mod);
           cout << ((mul(mul(b, b + 1, mod),ni,mod)- mul(mul(a, a + 1, mod), ni, mod)) % mod + mod) % mod << endl;
           continue;
       }
       LL ni = quick(n - 1, mod - 2, mod);
       LL anb = quick(n, bb, mod),ana=quick(n,aa,mod);
       anb = mul(b, anb, mod) - mul(ni,anb-1,mod);
       anb = (anb+mod) % mod;
       anb = mul(anb, ni, mod);
       ana = mul(a, ana, mod) - mul(ni, ana - 1, mod);
       ana = (ana % mod + mod) % mod;
       ana = mul(ana, ni, mod);
       cout <<((anb - ana)%mod+mod)%mod << endl;
   }
   return 0;
}