Codeforces Round #332 (Div. 2)

A. Patrick and Shopping

【题意】 Patrick 从家出发去超市，超市A离家d1，超市B离家d2，超市A,B相距d3， Patrick 要去两个超市并返回家。求最短距离。

【分析】一共四种情况，算出最小值

【代码】

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=0x3fffffff;
int main (void)
{
    int d1,d2,d3,t[4];
    scanf("%d%d%d",&d1,&d2,&d3);
    t[0]=d1+d2+d3;
    t[1]=2*d1+2*d3;
    t[2]=2*d1+2*d2;
    t[3]=2*d2+2*d3;
   sort(t,t+4);
   printf("%d
",t[0]);
}

B. Spongebob and Joke

【题意】给定长度为n的数组f和长度为m的数组b，问能否找到长度为m的数组a使得b[i]=f[a[i]]，若数组a不存在，输出impossible，若不唯一，输出Ambiguity，若唯一，输出impossible及数组a。

【分析】注意先判断是否是impossible再判断Ambiguity

【代码】

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=100050;
int b[maxn];
int a[maxn];
int cnt[maxn];
int sf[maxn];
int main (void)
{
    int m,n,temp, flag1=0,flag2=0;
    scanf("%d%d",&n,&m);
    fill(cnt,cnt+maxn,0);
    for(int i=0;i<n;i++)
      {
          scanf("%d",&temp);
          cnt[temp]++;
          sf[temp]=i+1;
      }
    for(int i=0;i<m;i++)  scanf("%d",&b[i]);
    for(int i=0;i<m;i++)
    {
        if(cnt[b[i]]==0)flag1=1;
        if(cnt[b[i]]>1) flag2=2;
        a[i]=sf[b[i]];
    }
    if(flag1==1)  printf("Impossible
");
    else  if(flag2==2)  printf("Ambiguity
");
    else
    {
        printf("Possible
");
        printf("%d",a[0]);
        for(int i=1;i<m;i++)   printf(" %d",a[i]);
    }
    return 0;
}

C. Day at the Beach

【题意】给定一系列数，对他们进行分块排序，使得最终整个数组为有序的，问最多可以分多少块。

【分析】可以直接判断该数前面的最大值是否小于后面的最小值，如果满足则可以将他单独划为一个block，否则将并到前面的block中。

Let’s take a minute to see how the best answer should look like. Let Hi be a sorted sequence of hi. Let E — set of indices of the last elements of each block. Then e E, first e sorted elements of sequence hi are equal to the first e elements of the sequence Hj. So, it is not difficult to notice that the size of E is the answer for the problem. Firstly, we need to calculate two arrays: prefmax and suffmin, where prefmaxi — maximum between a1, a2, …, ai, and suffmini — minimum between ai, ai + 1, …, an. If you want to get the answer, just calculate the number of indices i that prefmaxi  ≤  suffmini + 1.
Time: O(N)

【代码】

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
const int maxn=100050;
const int INF=0x3fffffff;
int h[maxn],suffMax[maxn],suffMin[maxn];
int main (void)
{
    int num,total=0;
    scanf("%d",&num);
    for(int i=1;i<=num;i++) scanf("%d",&h[i]);
    suffMin[num+1]=INF;
    suffMax[0]=-INF;
    for(int i=1;i<=num;i++)  suffMax[i]=max(suffMax[i-1],h[i]);
    for(int i=num;i>=1;i--)  suffMin[i]=min(suffMin[i+1],h[i]);
    for(int i=1;i<=num;i++)
    {
       if(suffMin[i+1]>=suffMax[i])
            total++;
    }
   printf("%d",total);
    return 0;
}

D. Spongebob and Squares

【题意】给定正方形个数n，求满足内部正方形总和为n的长方形个数，并输出每组长和宽

【分析】不难得出公式。悲哀的是明明公式已经找到可是写了半天就是wa。注意上限！！！！

设n<=m,则对于i在[1,n]，对(n-i+1)*(m-i+1)求和，化简之后为m*n*(n+1)/2-(n*n*n-n)/6

【代码】

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
const int maxn=1000000LL;
const int INF=0x3fffffff;
typedef long long ll;
vector<pair<ll,ll> > v;
int main (void)
{
    ll num;
    scanf("%I64d",&num);
    for(ll n=1;2*n*n*n+3*n*n+n<=7*num||n<=maxn;n++)
    {
       ll x1 = (n+1)*n/2ll;
       ll x2= (n*n*n-n)/6ll;
       ll temp=num+x2;
       if(temp%x1==0)
       {
              ll m=temp/x1;
              if(n>m) continue;
              v.push_back(make_pair(m,n));
              if(n!=m) v.push_back(make_pair(n,m));
       }
    }
    sort(v.begin(),v.end());
    printf("%d
",v.size());
    for(int i=0;i<v.size();i++)  printf("%I64d %I64d
",v[i].first,v[i].second);
    return 0;
}

本来题写的就慢，代码还断断续续改了好几天，，，