请求支援

01234567899870z @ 2017-11-04 19:43:08

by Zhao001 @ 2017-11-04 19:47:15

自己理解#include <iostream>

#include <cmath>
#include <cstring>
#include <cstdio>
using namespace std;
int s[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};
int main()
{
//    freopen("date.in","r",stdin);
//    freopen("date.out","w",stdout);
    int m,n,k=0,a=0,b=0;
    scanf("%d %d",&m,&n);
    if(m==n)
    {
        char x[10];
        int k=0;
        sprintf(x,"%d",m);
        for(int i=0;i<4;i++)
        {
            if(x[i]==x[7-i])
                k++;
        }
        if(k==4)
            cout<<'1';
        else
            cout<<'0';
        return 0;        
    }
    for(int i=1;i<=12;i++)
    {
        for(int j=1;j<=s[i];j++)
        {
            a=(j%10)*1000+(j/10)*100+(i%10)*10+(i/10);
            b=a*10000+i*100+j;
            if(b>=m&&b<=n)
                k++;              
        }
    }
    cout<<k;
    return 0;
}

by Zhao001 @ 2017-11-04 19:48:19

源自一位大牛的思路

by wenzhang @ 2017-12-10 17:38:57

#include<bits/stdc++.h>
using namespace std;
int hw(int x) {
    int y=0;
    while(x) {
        y=y*10+x%10;
        x/=10;
    }
    return y;
}
int pd(int a) {
    int r=a%100,y=a/100;
    if(y<=0||y>12||r<=0||r>31)return 0;
    if(y==4||y==6||y==9||y==11&&r>30)return 0;
    if(y==2&&r>29)return 0;
    return 1;
}
int i,a,b,n,m,ans,a1,b1;
int main() {
    scanf("%d%d",&a,&b);
    n=a/10000;
    m=b/10000;
    a1=a%10000;
    b1=b%10000;
    if(hw(n)>=a1&&pd(hw(n)))ans++;
    if(hw(m)<=b1&&pd(hw(m))&&n!=m)ans++;
    for(i=n+1; i<m; i++)
        if(pd(hw(i)))ans++;
    printf("%d\n",ans);
    return 0;
}70

by wenzhang @ 2017-12-10 17:39:30

??