Home Web Front-end HTML Tutorial cf#257(Div.2) B. Jzzhu and Sequences_html/css_WEB-ITnose

cf#257(Div.2) B. Jzzhu and Sequences_html/css_WEB-ITnose

Jun 24, 2016 pm 12:01 PM

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

2 33
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output

input

0 -12
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output

1000000006
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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:给你一个递推式,f2=f1 f3-->变形为:f[i]=f[i-1]-f[i-2];输入f[1]和f[2];求第n个数的值;如果你想打表的话劝你放弃吧!肯定会超时的,还记得高中的等差序列吗!对这绝对是一个周期序列!而且周期为6;

#include<stdio.h>int main(){   __int64 a[10],x,y,n;    int i;    scanf("%I64d%I64d",&x,&y);    a[1]=x;a[2]=y;    for(i=3;i<=6;i++)        a[i]=a[i-1]-a[i-2];    a[0]=a[6];    scanf("%I64d",&n);    n=n%6;    while(1)    {        if(a[n]<0)            a[n]+=1000000007;        else break;    }    printf("%I64d\n",a[n]%1000000007);    return 0;}
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