Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences
2014-07-20 01:03
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B. Jzzhu and Sequences
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Jzzhu has invented a kind of sequences, they meet the following property:
You are given x and y, please calculate fn modulo 1000000007 (109 + 7).
Input
The first line contains two integers x and y (|x|, |y| ≤ 109).
The second line contains a single integer n (1 ≤ n ≤ 2·109).
Output
Output a single integer representing fn modulo 1000000007 (109 + 7).
Sample test(s)
input
output
input
output
Note
In the first sample, f2 = f1 + f3, 3 = 2 + f3, f3 = 1.
In the second sample, f2 = - 1; - 1 modulo (109 + 7) equals (109 + 6).
题意:题意很简单,就是给定一个项数的计算方法,f [ i ] = f [ i-1 ] +f [ i + 1];
同时给定了f [ 1 ] 和 f [ 2 ];
让我们求f [ n ] mod 10^9+7 的值,其中如果f [ n ] 小于0 则对 算模n加的值,即 (10^9+7+f [ n ])mod(10^9 + 7);
好了,题目大意已知,接下来就是计算的问题了;
首先,如果说不是看到N的取值为 2 * 10^ 9 的话我是果断会暴力解决的。- -!
所以,常规的暴力方法就直接PASS掉吧。。。
思路: 既然暴力不行了,那就只能从项数公式入手解决了。。
仔细观察后咱会发现。。。。
如果把公式 f [ i ] = f [ i-1 ] +f [ i + 1];变为 f [ i ] = f [ i- 1 ]+f [ i - 2 ];的话。。。
得:
可以看出,它的是以7为循环的数,所以就可以很容易的写出代码了,代码如下:
如发现bug,欢迎指出,万分感谢!!
以上。
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Jzzhu has invented a kind of sequences, they meet the following property:
You are given x and y, please calculate fn modulo 1000000007 (109 + 7).
Input
The first line contains two integers x and y (|x|, |y| ≤ 109).
The second line contains a single integer n (1 ≤ n ≤ 2·109).
Output
Output a single integer representing fn modulo 1000000007 (109 + 7).
Sample test(s)
input
2 3 3
output
1
input
0 -12
output
1000000006
Note
In the first sample, f2 = f1 + f3, 3 = 2 + f3, f3 = 1.
In the second sample, f2 = - 1; - 1 modulo (109 + 7) equals (109 + 6).
题意:题意很简单,就是给定一个项数的计算方法,f [ i ] = f [ i-1 ] +f [ i + 1];
同时给定了f [ 1 ] 和 f [ 2 ];
让我们求f [ n ] mod 10^9+7 的值,其中如果f [ n ] 小于0 则对 算模n加的值,即 (10^9+7+f [ n ])mod(10^9 + 7);
好了,题目大意已知,接下来就是计算的问题了;
首先,如果说不是看到N的取值为 2 * 10^ 9 的话我是果断会暴力解决的。- -!
所以,常规的暴力方法就直接PASS掉吧。。。
思路: 既然暴力不行了,那就只能从项数公式入手解决了。。
仔细观察后咱会发现。。。。
如果把公式 f [ i ] = f [ i-1 ] +f [ i + 1];变为 f [ i ] = f [ i- 1 ]+f [ i - 2 ];的话。。。
得:
f[1]=f1; f[2]=f2; f[3]=f[2]-f[1]=f2-f1; f[4]=f[3]-f[2]=-f1; f[5]=f[4]-f[3]=-f2; f[6]=f[5]-f[4]=-f2+f1; f[7]=f[6]-f[5]=f1;
可以看出,它的是以7为循环的数,所以就可以很容易的写出代码了,代码如下:
#include <iostream> using namespace std; const int maxx=1000000007; int main( ) { int f[7]; int f1,f2; cin>>f1>>f2; f[1]=f1; f[2]=f2; f[3]=f2-f1; f[4]=-f1; f[5]=-f2; f[0]=-f2+f1; int n; cin>>n; n=n%6; f =f %maxx; if(f <0) { cout<<maxx+f <<endl; } else { cout<<f <<endl; } return 0; }
如发现bug,欢迎指出,万分感谢!!
以上。
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