标签：... ace ios depend ready using input start include
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6740 | Accepted: 3276 |
Description
Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.
Input
The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.
Output
For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.
Sample Input
2 1 2
Sample Output
2 6
#include <iostream> using namespace std; const int mod=10007; int pow(int a,int b){//quilck_^ int ans=1,p=a;//make ans as 1,p as a while(b){//if b isnot 0 ,just go on if(b%2==1)ans=(ans*p)%mod;//if b is an odd number , //just let the ans multiply the given base_number and % the given mod //!!if you don‘t %,maybe it at first larger than the range of int b=b/2;p=(p*p)%mod;//let the index been divided by 2 //and renew the p ,and let it % the given mod } //if b is 0 , just means has ^ already ,just break return ans; //then return the ans } int main(){int t; scanf("%d",&t);//given case_number while(t--){int n;scanf("%d",&n);//depend on the case_number printf("%d\n",(pow(4,n-1)+pow(2,n-1))%mod); //after matrix eigenvalues }return 0; } //4^n=(2+1+1)^4，ans=sum[i=0...n] //(c(n,i)2^i*sum[k=n-i and it is an even_number,t=0,2,...k](c(k,t) ); //at last work out 4^(n-1)+2^(n-1)。 //blocks
program is above, the program is the best language
标签：... ace ios depend ready using input start include
原文地址：http://www.cnblogs.com/muzu/p/7145094.html
