2017年ICPC中国大陆区域赛真题（上）

2017年ICPC中国大陆区域赛真题（上）

A - Cats and Fish

There are many homeless cats in PKU campus. They are all happy because the students in the cat club of PKU take good care of them. Li lei is one of the members of the cat club. He loves those cats very much. Last week, he won a scholarship and he wanted to share his pleasure with cats. So he bought some really tasty fish to feed them, and watched them eating with great pleasure. At the same time, he found an interesting question:

There are m fish and n cats, and it takes c_i minutes for the i^th cat to eat out one fish. A cat starts to eat another fish (if it can get one) immediately after it has finished one fish. A cat never shares its fish with other cats. When there are not enough fish left, the cat which eats quicker has higher priority to get a fish than the cat which eats slower. All cats start eating at the same time. Li Lei wanted to know, after x minutes, how many fish would be left.

Input

There are no more than 20 test cases.

For each test case:

The first line contains 3 integers: above mentioned m, n and x (0 < m <= 5000, 1 <= n <= 100, 0 <= x <= 1000).

The second line contains n integers c₁,c₂ … c_n,  c_i means that it takes the ith cat ci minutes to eat out a fish ( 1<= c_i <= 2000).

Output

For each test case, print 2 integers p and q, meaning that there are p complete fish(whole fish) and q incomplete fish left after x minutes.

Sample Input

2 1 1
1
8 3 5
1 3 4
4 5 1
5 4 3 2 1

Sample Output

1 0
0 1
0 3

#include <bits/stdc++.h>

using namespace std;
int c[10000],cnt,b[10000],num;
int main()
{
    int m,n,x;
    while (~scanf("%d%d%d",&m,&n,&x))
    {
        for (int i=1; i<=n; i++)
        {
            scanf("%d",&c[i]);
        }
        memset(b,0,sizeof(b));
        cnt=m;
        num=0;
        sort(c+1,c+n+1);
        for (int i=1; i<=x; i++)
        {
            for (int j=1; j<=n; j++)
            {
                if (b[j]==0)
                {
                    cnt--;
                }
                if (i%c[j]==0)
                {
                    b[j]=0;
                    num++;
                }
                else
                {
                    b[j]=1;
                }
                if (cnt==0)
                {
                    break;
                }
            }
            if (cnt==0)
            {
                break;
            }
        }
        int ans=0;
        for (int i=1; i<=n; i++)
        {
            if (b[i]==1)
            {
                ans++;
            }
        }
        printf("%d %d
",m-num-ans,ans);
    }
}

　　

B - Secret Poems

#include<bits/stdc++.h>

using namespace std;

int n,a[100][100];
char c[100][100],cc[10000];
int b[1000][100];
int dx[]= {0,1,0,-1};
int dy[]= {1,0,-1,0};
void snakeLikeMat(int n)
{
    int num = 1;
    int total = n*n;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j <= i; j++)
        {
            if (i % 2 == 0)
            {
                a[i-j][j] = num++;
            }
            else
            {
                a[j][i-j] = num++;
            }
        }
    }
    for (int i = 0; i < n-1; i++)
    {
        for (int j = 0; j <= i; j++)
        {
            if (i%2==0)
            {
                a[n - 1 - i+ j][n - 1 - j] = total--;
            }
            else
            {
                a[n - 1 - j][n - 1 - i + j] = total--;
            }
        }
    }
}

int main()
{
    while (~scanf("%d",&n))
    {
        memset(b,0,sizeof(b));
        snakeLikeMat(n);
        for (int i=0; i<n; i++)
        {
            scanf("%s",c[i]);
        }
        for (int i=0; i<n; i++)
        {
            for (int j=0; j<n; j++)
            {
                cc[a[i][j]-1]=c[i][j];
                c[i][j]=' ';
            }
        }
        int x=0,y=-1,k=0,xx,yy;
        for (int i=0; i<n*n; i++)
        {
            xx=x+dx[k%4];
            yy=y+dy[k%4];
            if (xx>=n||xx<0||yy<0||yy>=n||b[xx][yy])
            {
                k++;
                xx=x+dx[k%4];
                yy=y+dy[k%4];
            }
            c[xx][yy]=cc[i];
            b[xx][yy]=1;
            x=xx;
            y=yy;
        }
        for (int i=0; i<n; i++)
        {
            printf("%s
",c[i]);
        }
    }
}

　　

C - Pangu and Stones

题意：

n个石子堆排成一排，每次可以将连续的最少L堆，最多R堆石子合并在一起，消耗的代价为要合并的石子总数

求合并成1堆的最小代价，如果无法做到输出0

思路：

dp[i][j][k]表示区间[i, j]分成k堆的最小代价，转移有

k=1时：

dp[i][j][1] = min(dp[i][p][x-1]+dp[p+1][j][1]+sum[i][j] (i<=p<=j-1；L<=x<=R) )

k>1时：

dp[i][j][k] = min(dp[i][p][k-1]+dp[p+1][j][1] ( i<=p<=j-1) )

#include<bits/stdc++.h>

using namespace std;
const int maxn=110;
int n,l,r,a[maxn],sum[maxn],dp[maxn][maxn][maxn];
int main()
{
    while (~scanf("%d%d%d",&n,&l,&r))
    {
        for (int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        memset(dp,0x3f3f3f,sizeof(dp));
        for (int i=1; i<=n; i++)
        {
            for (int j=i; j<=n; j++)
            {
                dp[i][j][j-i+1]=0;
            }
        }
        for (int p=1; p<n; p++)
        {
            for (int i=1; i+p<=n; i++)
            {
                for (int j=i; j<=i+p-1; j++)
                {
                    for (int k=l-1; k<=r-1; k++)
                    {
                        dp[i][i+p][1]=min(dp[i][i+p][1],dp[i][j][k]+dp[j+1][i+p][1]+sum[i+p]-sum[i-1]);
                    }
                }
                for (int j=2; j<=p; j++)
                {
                    for (int k=i; k<=i+p-1; k++)
                    {
                        dp[i][i+p][j]=min(dp[i][i+p][j],dp[i][k][j-1]+dp[k+1][i+p][1]);
                    }
                }
            }
        }
        if (dp[1][n][1]==0x3f3f3f3f)
        {
            printf("0
");
        }
        else
        {
            printf("%d
",dp[1][n][1]);
        }
    }
    return 0;
}

　　

Q - Puzzle Game