Codeforces Round #324 (Div. 2) B. Kolya and Tanya 思维题 数论

http://codeforces.com/contest/584/problem/B

题意：给你3n个点，均匀分散在一个圆的周围，每三个下标符合i,i+n,i+2n的点构成一个组合，

每个点可取a=1,2,3,三个数字，问给定了n后，求出有多少种圆中不存在任何一种ai+a(i+n)+a(i+2n)=6的

组合数；

错因分析：核心思想分析的出来，就是不会处理计算；

分析：若 a>b，则(a-b)%m==（a%m-b%m+m）注意中间要加个m，因为可能a%m<b%m；

AC代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        ll s = 27, w = 7;
        for (int i = 1; i <= n - 1; i++)
        {
            s = (s * 27) % (1000000000 + 7);
            w = (w * 7) %  (1000000000 + 7);
        }
        printf("%lld\n", ( s-w+1000000000+7)%(1000000000+7));
    }
    return 0;
}

WA代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        ll s = 27, w = 7;
        for (int i = 1; i <= n - 1; i++)
        {
            s = (s * 27) % (1000000000 + 7);
            w = (w * 7) % (1000000000 + 7);
        }
        printf("%lld\n", s-w);
    }
    return 0;
}