版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/liufengwei1/article/details/82179876
昨天做了一套2016湖南省赛的2016,签到题不会做,队友想出来的,于是看了题解,来做这道今年湘潭比赛的题,然后发现那个方法在这个1e4组数据里面会T,然而因为2018只有2个质因子,所以可以直接容斥。我们先算出a-b之间的被2整除但不被2018整除的数字,再算出c-d之间被2整除但不被2018整除的数字,然后乘一蛤加到答案里,接着a-b,c-d交换,再乘加到答案里,接着考虑2018,找出a-b区间中2018的倍数*c-d区间长度,反之亦然,但是这样会重复计算a-b中2018倍数*c-d中2018倍数,减掉就好了
#include<cstdio>
#include<cstring>
long long a,b,c,d,ans;
int main()
{
long long xnum,ynum;
while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&d))
{
ans=0;
xnum=b/2-(a-1)/2-b/2018+(a-1)/2018;
ynum=d/1009-(c-1)/1009-d/2018+(c-1)/2018;
ans+=xnum*ynum;
xnum=b/1009-(a-1)/1009-b/2018+(a-1)/2018;
ynum=d/2-(c-1)/2-d/2018+(c-1)/2018;
ans+=xnum*ynum;
xnum=b/2018-(a-1)/2018;ynum=d/2018-(c-1)/2018;
ans+=xnum*(d-c+1);
ans+=ynum*(b-a+1);
ans-=xnum*ynum;
printf("%lld\n",ans);
}
return 0;
}