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Description
The tuk-tuk (????), also known as "auto rickshaw," is a popular form of transportation in Thailand. In order to distinguish tuk-tuks from the other types of vehicles, the Phuket administration decided to create a new license plate system for them. Phuket‘s tuk-tuk fleet has been growing rapidly, mostly due to tourism, one of the most important economic activities in the province. The administration plans the new system to meet the demand for distinct license plates for the next 42 years.
A license plate system is defined by two numbers, C and D. A license plate in this system is a string formed by C consonants followed by D digits. A license plate cannot be empty (no consonant and no digit).
In the Thai alphabet there are 44 consonants and 10 digits. However, since the symbols of some consonants are too much alike, the administration decided to use only 26 consonants, whose symbols were considered sufficiently distinguishable.
To guarantee an appropriate supply of tuk-tuks for the contestants in the ICPC World Finals in 2016, Phuket‘s administration would like to know the number of distinct license plates it is possible to generate with a given license plate system.
Input
The first line has a single integer T, the number of test cases.
Each test case consists of a single line containing two integers C and D, the number of consonants and digits, respectively, in a license plate system.
Limits
Output
For each test case, print a line with the amount of distinct license plates that can be generated by the corresponding system. The answer is guaranteed to be smaller than 231.
Sample Input
3
0 6
2 4
0 0
1000000
6760000
0
套模板速度一发a
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w",stdout);
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;
const double PI = acos(-1);
LL fastpow(LL a,LL b,LL c){
LL res = a;
while(c){
if(c & 1) res = (res * b);
b = (b * b);
c >>= 1;
}
return res;
}
int main()
{
//FIN
int T;
scanf("%d", &T);
while(T--){
LL a, b;
scanf("%I64d%I64d", &a, &b);
LL ans = 1;
ans = fastpow(1, 26, a) * fastpow(1, 10, b);
if(a == b && a == 0) ans = 0;
printf("%I64d\n", ans);
}
}
CodeForces Gym 101047L Putting plates on the tuk-tuks 快速幂
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原文地址：http://www.cnblogs.com/Hyouka/p/5802039.html
