zoukankan      html  css  js  c++  java
  • Codeforces Round #257 (Div. 2 ) B. Jzzhu and Sequences

    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 3
    3
    Output
    1
    Input
    0 -1
    2
    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).

     1 #include<iostream>
     2 #include<string.h>
     3 #include<stdio.h>
     4 #include<ctype.h>
     5 #include<algorithm>
     6 #include<stack>
     7 #include<queue>
     8 #include<set>
     9 #include<math.h>
    10 #include<vector>
    11 #include<map>
    12 #include<deque>
    13 #include<list>
    14 using namespace std;
    15 #define M 1000000007
    16 int main()
    17 {
    18     __int64 n,a[6];
    19     scanf("%I64d%I64d%I64d", &a[0], &a[1], &n);
    20     a[0]=a[0]%M;
    21     a[1]=a[1]%M;
    22     a[2]=(a[1]-a[0])%M;
    23     a[3]=(-a[0])%M;
    24     a[4]=(-a[1])%M;
    25     a[5]=(a[0]-a[1])%M;
    26     printf("%I64d
    ", a[(n-1)%6]>=0? a[(n-1)%6] : a[(n-1)%6]+M);
    27     
    28     
    29     return 0;
    30 }
    View Code
  • 相关阅读:
    C++ 什么是多态
    *和&的使用
    静态链接库与动态链接库
    利尔达CC3200模块烧写程序笔记
    创龙TMS320C6748开发找不到 tl.dsp.evm6748的问题研究
    RTSC和XDCTool的理解
    创龙DSP6748开发板SYS/BIOS的LED闪烁-第2篇
    Coap协议学习笔记-第一篇
    linux进程的学习笔记(未完)
    创龙DSP6748开发板LED闪烁-第一篇
  • 原文地址:https://www.cnblogs.com/qscqesze/p/3856487.html
Copyright © 2011-2022 走看看