思路:
这题写的真是心态要炸啊,数学渣渣一枚,想的太多脑袋就是浆糊了,不过还好A啦,总算没有丢人…
2018的因子有:1,2,1009,2018,分类讨论即可:
1. [a,b]中2018的倍数,[c,d]为任意数
2. [c,d]中2018的倍数,[a,b]为任意数
3. [a,b]中2018的倍数且[c,d]中2018的倍数(为了1,2情况去重)
4. [a,b]中1009的奇数倍(偶数倍同1有重叠),[d,c]中2的倍数且不是2018的倍数
5. [c,d]中1009的奇数倍(偶数倍同2有重叠),[a,b]中2的倍数且不是2018的倍数
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
ll a,b,c,d;
ll f(ll x){
x/=1009;
if(x%2)return x/2+1;
return x/2;
}
int main()
{
while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&d)){
ll sum=0;
sum+=(b/2018-(a-1)/2018)*(d-c+1); // 1
sum+=(d/2018-(c-1)/2018)*(b-a+1); // 2
sum-=(b/2018-(a-1)/2018)*(d/2018-(c-1)/2018); // 3
sum+=(f(b)-f(a-1))*(d/2-(c-1)/2-(d/2018-(c-1)/2018)); // 4
sum+=(f(d)-f(c-1))*(b/2-(a-1)/2-(b/2018-(a-1)/2018)); //5
printf("%lld\n",sum);
}
return 0;
}