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Light OJ 1006 - Hex-a-bonacci

发布时间：2024/8/24 编程问答 41 如意码农
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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]);

     }

 }

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