Light OJ 1006

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 == 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;
    return( fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6) );
}
int main() {
    int n, caseno = 0, 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
", ++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.

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=1,i,n;

    scanf("%d", &T);

    while(T--)
    {
        for(i=0;i<6;i++)
        {
            scanf("%d", &a[i]);
            a[i]%=10000007;///取余是必须滴
        }

        scanf("%d", &n);

        for(i=6;i<=n;i++)
        {
            a[i]=(a[i-1]+a[i-2]+a[i-3]+a[i-4]+a[i-5]+a[i-6])%10000007;
        }

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