Codeforces Round #257 (Div. 2/B)/Codeforces450B_Jzzhu and Sequences

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B解题报告

算是规律题吧，，，x y z -x -y -z

注意的是如果数是小于0,要先对负数求模再加模再求模,不能直接加mod，可能还是负数

给我的戳代码跪了，，，

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
long long x,y,z;
long long n;
int main()
{
    cin>>x>>y;
    cin>>n;
    z=y-x;
    long long t;
    t=(n-1)/3;
    if(t%2==0)
    {
        n%=3;
        if(n==0)
        {
            if(z>=0)
            cout<<z%1000000007;
            else cout<<(z%1000000007+1000000007)%1000000007;
        }
        else if(n==1)
        {
            if(x>=0)
            cout<<x%1000000007;
            else cout<<(x%1000000007+1000000007)%1000000007;
        }
        else
        {
            if(y>=0)
            cout<<y%1000000007;
            else cout<<(y%1000000007+1000000007)%1000000007;
        }
    }
    else
    {
        x=-x;
        y=-y;
        z=-z;
        n%=3;if(n==0)
        {
            if(z>=0)
            cout<<z%1000000007;
            else cout<<(z%1000000007+1000000007)%1000000007;
        }
        else if(n==1)
        {
            if(x>=0)
            cout<<x%1000000007;
            else cout<<(x%1000000007+1000000007)%1000000007;
        }
        else
        {
            if(y>=0)
            cout<<y%1000000007;
            else cout<<(y%1000000007+1000000007)%1000000007;
        }
    }
    return 0;
}

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).

2 3
3

1

0 -1
2

1000000006

Codeforces Round #257 (Div. 2/B)/Codeforces450B_Jzzhu and Sequences,布布扣,bubuko.com

Codeforces Round #257 (Div. 2/B)/Codeforces450B_Jzzhu and Sequences

标签：style   http   color   os   width   io   

原文地址：http://blog.csdn.net/juncoder/article/details/37974179