HDOJ 6768 The Oculus

http://acm.hdu.edu.cn/showproblem.php?pid=6768

题意：

一个正整数 表示为 斐波那契数列数列。

存在当前第i个斐波那契数 bi 为1，否则为0。

给三个斐波那契数列表示A B C，问A * B与C修改哪一位之后 的b值相等。

思路：

哈希 求模 再得差值

找到差值在哪一位

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=2e5+10;
const int mod=1e9+7;

ll a[maxn] = {0};
ll b[maxn] = {0};
int n[maxn];
int main()
{
 	int T = 0;
    a[1] = 1;
	b[1] = 1;
	a[2] = 2;
	b[2] = 2;
    for(int i = 3;i<maxn;i++)
	{
		int value = 0;
	 	cin >> T;
	 	a[i] = (a[i-1] + a[i-2]) % mod;
		b[i] = (b[i-1] + b[i-2]) % mod;
		int p = 0;
		while(T--)
		{
		 	ll aa = 0 , aa1 = 0;
		    ll bb = 0 , bb1 = 0;
        	int k = 0;
        	cin >> k;
        	for(int j = 1;j<=k;j++)
			{
			 	cin >> value;
            	if(value != 0)
					{
					 	aa += a[j];
						aa %= mod;
                		aa1 += b[j];
						aa1 %= mod;
   					}
		    }
        	cin >> k;
        	value = 0;
        	for(int j = 1;j<=k;j++)
			{
        	    cin >> value;
        		    if(value != 0)
					{
            		    bb += a[j];
						bb %= mod;
		    			bb1 += b[j];
						bb1 %= mod;
   					}
        	}
        	ll res_1 = aa * bb % mod;
			ll res_2 = aa1 * bb1 % mod;
        	cin >> k;
        	ll add1 = 0;
			ll add2 = 0;
        	for(int i = 1;i <= k;i++)
			{
        	    cin >> n[i];
        		if(n[i] != 0)
				{
                	add1 = (add1 + a[i]) % mod;
                	add2 = (add2 + b[i]) % mod;
            	}
        	}
        	for(int i = 1;i<=k;i++)
			{
        	    if(n[i] == 0)
				{
            	    ll val1 = (add1 + a[i]) % mod;
            	    ll val2 = (add2 + b[i]) % mod;
    				if(val1 == add1 && val2 == add2)
					{
                	    p = i;
							break;
	    			}
      			}
        	}
		cout << p;
    }
    return 0;
}