cxm1024 @ 2019-11-02 08:28:34

#include<iostream>
#include<cmath>
using namespace std;
int daynum[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
void jia(int *x)
{
    (*x)++;
    int nian=(*x)/10000,yue=(*x)/100%10000,ri=(*x)%100;
    if(nian%400==0)daynum[2]=29;
    else if(nian%100==0)daynum[2]=28;
    else if(nian%4==0)daynum[2]=29;
    else daynum[2]=28;
    if(ri>daynum[yue])(*x)-=daynum[yue],(*x)+=100;
    if(yue>12)(*x)-=1200,(*x)+=10000;
}
bool pd(int x)
{
    bool b=1;
    for(int i=1;i<=4;i++)
        if(x%int(pow(10,i))/int(pow(10,i-1))!=x%int(pow(10,8-i+1))/int(pow(10,8-i)))
        {
            b=0;break;
        }
    return b;
}
int main()
{
    int l,r,ans=0;
    cin>>l>>r;
    for(;l<=r;jia(&l))
        if(pd(l))
            ans++;
    cout<<ans<<endl;
    return 0;
}

by cxm1024 @ 2019-11-02 08:32:28

更易理解版

#include<iostream>
#include<cmath>
using namespace std;
int daynum[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
bool pd(int x)
{
    bool b=1;
    for(int i=1;i<=4;i++)
        if(x%int(pow(10,i))/int(pow(10,i-1))!=x%int(pow(10,8-i+1))/int(pow(10,8-i)))
        {
            b=0;break;
        }
    return b;
}
int main()
{
    int l,r,ans=0;
    cin>>l>>r;
    while(l<=r)
    {
        if(pd(l))ans++;
        l++;
        int nian=l/10000,yue=l/100%10000,ri=l%100;
        if(nian%400==0)daynum[2]=29;
        else if(nian%100==0)daynum[2]=28;
        else if(nian%4==0)daynum[2]=29;
        else daynum[2]=28;
        if(ri>daynum[yue])l-=daynum[yue],l+=100;
        if(yue>12)l-=1200,l+=10000;
    }
    cout<<ans<<endl;
    return 0;
}

by Junthe @ 2019-11-02 08:40:01

哪题

Which problem

by sandwich @ 2019-11-02 08:49:32

好像没什么毛病哦

by touxi @ 2020-06-11 19:26:55

daynum[]={0,31,28,31,30,31,30,31,31,30,31,30,31};

by touxi @ 2020-06-11 19:27:14

daynum[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};