Java 解法, 执行用时: 43ms, 内存消耗: 10584K, 提交时间: 2022-08-10 14:01:50

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int k=sc.nextInt();
        int x=sc.nextInt();
        int y=sc.nextInt();
        int len,count=0;
        len=y-x+1;
        for (int i = 1,t=0; i <=k ; i+=2,t++) {
            if ((t&1)==1)
            {
                if ((x&1)==1)
                {
                    count+=len/2;
                }
                else
                {
                    count+=(len+1)/2;
                }
            }
            else
        
                count+=len;
            count%=1000000007;
        
        }
        System.out.println(count);
    }
}

C++14(g++5.4) 解法, 执行用时: 13ms, 内存消耗: 540K, 提交时间: 2019-12-17 20:14:53

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef long long LL;
const int MAXN = 1005;
const int MOD = 1e9+7;
//x%4==1 --> all 1
//x%4==3 --> 01
signed main(){
    //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    int k,x,y; cin>>k>>x>>y;
    int res=0,t1=k/4,t2=k%4;
    int a=t1+(t2>=1),b=t1+(t2>=3);
    if(x&1) res = (a*(y-x+1)+b*((y-x+1)/2))%MOD;
    else res = (a*(y-x+1)+b*((y-x)/2+1))%MOD;
    cout<<res<<endl;
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 5ms, 内存消耗: 504K, 提交时间: 2019-12-19 00:13:07

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod = 1e9+7;
int main()
{
    ll k, x, y; cin >> k >> x >> y;
    k = k - k / 2;
    ll a = k / 2;
    cout << (a * (y / 2 - (x - 1) / 2) + (k - a) * (y - x + 1)) % mod << "\n";
    return 0;
}