Light OJ 1006 Hex-a-bonacci（記憶化搜尋優化）
Given a code (not optimized), and necessary inputs, you have to find the output of the code for the inputs. The code is as follows:
int a, b, c, d, e, f;
int fn( int n ) {
if( n == 0 ) return a
if( n == 1 ) return b;
if( n == 2 ) return c;
if( n == 3 ) return d;
if( n == 4 ) return
if( n == 5 ) return f;
return( fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6) );
}
int main() {
int n, caseno
scanf("%d", &cases);
while( cases-- ) {
scanf("%d %d %d %d %d %d %d", &a, &b, &c, &d, &e, &f, &n);
printf("Case %d: %d\n", ++caseno, fn(n) % 10000007);
}
return 0;
}
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case contains seven integers, a, b, c, d, e, f and n. All integers will be non-negative and 0 ≤ n ≤ 10000 and the each of the others will be fit into a 32-bit integer.
Output
For each case, print the output of the given code. The given code may have integer overflow problem in the compiler, so be careful.
Sample Input
5
0 1 2 3 4 5 20
3 2 1 5 0 1 9
4 12 9 4 5 6 15
9 8 7 6 5 4 3
3 4 3 2 54 5 4
Sample Output
Case 1: 216339
Case 2: 79
Case 3: 16636
Case 4: 6
Case 5: 54
主要是對已有的程式碼進行優化，把計算過的值直接返回，用記憶化搜尋進行：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string.h>
using namespace std;
int a, b, c, d, e, f;
int dp[10005];
int fn( int n ) {
if(dp[n]!=0) return dp[n];//優化
if( n == 0 ) return a;
if( n == 1 ) return b;
if( n == 2 ) return c;
if( n == 3 ) return d;
if( n == 4 ) return e;
if( n == 5 ) return f;
dp[n]=( fn(n-1) % 10000007 + fn(n-2) % 10000007 + fn(n-3) % 10000007 + fn(n-4) % 10000007 + fn(n-5) % 10000007 + fn(n-6) % 10000007) % 10000007;
return dp[n];
}
int main() {
int n, caseno = 0, cases;
scanf("%d", &cases);
while( cases-- )
{
memset(dp,0,sizeof(dp));
scanf("%d %d %d %d %d %d %d", &a, &b, &c, &d, &e, &f, &n);
printf("Case %d: %d\n", ++caseno, fn(n) % 10000007);
}
return 0;
}