Light OJ 1006 - Hex-a-bonacci

题目链接：http://acm.hust.edu.cn/vjudge/contest/121396#problem/G

http://lightoj.com/volume_showproblem.php?problem=1006

密码：acm

Description

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 ==  ) return a;
    if( n ==  ) return b;
    if( n ==  ) return c;
    if( n ==  ) return d;
    if( n ==  ) return e;
    if( n ==  ) return f;
    return( fn(n-) + fn(n-) + fn(n-) + fn(n-) + fn(n-) + fn(n-) );
}
int main() {
    int n, caseno = , cases;
    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) % );
    }
    return ;
}

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.

分析：改善整数溢出情况

*：数组、求余

 #include<stdio.h>
 #include<math.h>
 #include<string.h>
 #include<stdlib.h>
 #include<algorithm>
 #include<iostream>

 using namespace std;

 #define N 11000

 int a[N];

 int main()
 {
     int T,k=,i,n;

     scanf("%d", &T);

     while(T--)
     {
         for(i=;i<;i++)
         {
             scanf("%d", &a[i]);
             a[i]%=;///取余是必须滴
         }

         scanf("%d", &n);

         for(i=;i<=n;i++)
         {
             a[i]=(a[i-]+a[i-]+a[i-]+a[i-]+a[i-]+a[i-])%;
         }

         printf("Case %d: %d\n", k++, a[n]);
     }
 }