2017"百度之星"程序设计大赛 - 初赛（A）

最新推荐文章于 2022-08-28 17:07:58 发布

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最新推荐文章于 2022-08-28 17:07:58 发布

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原文链接：http://www.cnblogs.com/sbfhy/p/7351236.html

本文解析了2017年百度之星程序设计大赛初赛中的三道题目，包括求n-1因子数量、日期计算及并查集应用。通过具体代码展示了如何使用C++解决这些问题。

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2017"百度之星"程序设计大赛 - 初赛（A）

hdu6108 求出 n-1 的因子个数即可

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 200005;

int main()
{
    int T;  scanf("%d", &T);
    int n;
    while(T--)
    {
        scanf("%d", &n);
        int ans=0;
        for(int i=1; i<(int)sqrt(n-1); ++i)
            if((n-1)%i==0) ans += 2;
        if((n-1)%(int)sqrt(n-1)==0) ++ans;
        printf("%d\n", ans);
    }

    return 0;
}

hdu6112 暴力模拟

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 200005;

int num[15][35];
int m[15]={0,31,28,31,30,31,30,31,31,30,31,30,31};
void getnum()
{
    int sum=0;
    rep(i,1,12) rep(j,1,m[i])
    {
        ++sum,  num[i][j]=sum;
    }
    num[2][29]=num[2][28];
}
int is(int year)
{
    return ( year%400==0 || (year%4==0 && year%100!=0) );
}
int isday(int year, int mon, int day)
{
    return num[mon][day] + (is(year) && (mon>2 || (mon==2 && day==29)));
}
int main()
{
    getnum();  
    int T;  scanf("%d", &T);
    while(T--)
    {
        int year, mon, day;
        scanf("%d-%d-%d", &year, &mon, &day);
        ll  cnt=0, cnt1;
        rep(i,2001,year-1) cnt += 365+is(i);
        cnt1 = cnt;
        cnt += isday(year,mon,day);
        int dat = cnt%7,  ans;  
        rep(i,year,10000)
        {
            cnt1 += 365+is(i);
            if(!(mon==2 && day==29) || is(i+1))
            {
                cnt1 += isday(i+1,mon,day);
                if(cnt1%7==dat) { ans=i+1; break; }
                cnt1 -= isday(i+1,mon,day);
            }
        }
        printf("%d\n", ans);
    }

    return 0;
}

hdu6113 并查集，找出连通快判断即可

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 110;

int id[N][N], fa[N*N], n, m, num0[N*N];
char s[N][N];
bool is(int x, int y)
{
    return x>0 && y>0 && x<=n && y<=m;
}
int Find(int x) {
    return fa[x]==x ? x : fa[x]=Find(fa[x]);
}
void Unite(int a, int b)
{
    int faa=Find(a), fab=Find(b);
    if(faa!=fab) fa[faa]=fa[fab];
}
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        int tot=0;
        rep(i,1,n)
        {
            scanf("%s", s[i]+1);
            rep(j,1,m) id[i][j]=++tot;
        }
        rep(i,1,tot) fa[i] = i;
        rep(i,1,n) rep(j,1,m)
        {
            if(is(i-1,j) && s[i-1][j]==s[i][j]) Unite(id[i-1][j], id[i][j]);
            if(is(i+1,j) && s[i+1][j]==s[i][j]) Unite(id[i+1][j], id[i][j]);
            if(is(i,j+1) && s[i][j+1]==s[i][j]) Unite(id[i][j+1], id[i][j]);
            if(is(i,j-1) && s[i][j-1]==s[i][j]) Unite(id[i][j-1], id[i][j]);
        }
        int  cnt=0;   mes(num0, 0);
        int cnt1=0;
        rep(i,1,n) rep(j,1,m)
        {
            if(s[i][j]=='1' && fa[id[i][j]]==id[i][j]) ++cnt;
            if(s[i][j]=='0' && fa[id[i][j]]==id[i][j]) ++num0[fa[id[i][j]]], ++cnt1;
        }
        if(cnt!=1) { puts("-1"); continue; }
        cnt=0;
        rep(ca,1,tot) if(num0[ca])
        {
            bool  flag=0;
            rep(i,1,n) {
                rep(j,1,m) if(s[i][j]=='0' && fa[id[i][j]]==ca) {
                    if(i==1 || i==n || j==1 || j==m) { flag=1;  break; }
                }
                if(flag) break;
            }
            if(flag==0) ++cnt;
        }
        if(cnt==1) puts("0");
        else if(cnt==0) puts("1");
        else puts("-1");
    }

    return 0;
}