Educational Codeforces Round 5(A,B,C,D)

tags: Codeforces

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果然对于我这样的弱鸡来说还是Educational敲着比较畅快。然而因为自己的脑残在D题折腾了一下午，，，一对括号引发的血案~~(╯﹏╰)b

A.Comparing Two Long Integers

解析

比较两个超级大的数字的大小，去掉前导0后，剩下的长度越长数字越大，一样长的时候自然是用strcmp()进行字符串比较啦。

代码

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cassert>#include <algorithm>using namespace std;#define MAXN    1000000+100#define MOD     1000000007char str1[MAXN], str2[MAXN];int main(){    scanf("%s%s",str1,str2);    int len1 = strlen(str1);    int len2 = strlen(str2);    int p1 = 0;    while (p1 < len1&&str1[p1] == '0')        p1++;    len1 -= p1;    int p2 = 0;    while (p2 < len2&&str2[p2] == '0')        p2++;    len2 -= p2;    int ans = 0;    if (len1 != len2)    {        if (len1 > len2)            ans = 1;        else            ans = -1;    }    else    {        ans = strcmp(&str1[p1],&str2[p2]);    }    if (ans == 0)        printf("=\n");    else if (ans == 1)        printf(">\n");    else if (ans == -1)        printf("<\n");    return 0;}

B.Dinner with Emma

解析

显而易见，就是找每行中的最小值，然后取其中的最大值即可。

代码

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cassert>#include <algorithm>using namespace std;#define MAXN    1000000+100#define MOD     1000000007int main(){    int m, n;    scanf("%d%d", &n, &m);    long long  maxc = 0;    for (int i = 0; i < n; i++)    {        long long k;        long long minc=1e10;        for (int j = 0; j < m; j++)        {            scanf("%lld",&k);            if (k < minc)                minc = k;        }        if (minc > maxc)            maxc = minc;    }    printf("%lld\n",maxc);    return 0;}

C.The Labyrinth

解析

先使用搜索预处理出每个’.’组成的连通分量的大小，并将连通分量编号，同时将每个‘.’所在连通分量的大小存在sz[MAXN][MAXN]数组中。

对于每个‘#’，统计和它相连的不同连通分量的大小，然后求和后取模即可得到该点的值。

代码

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cassert>#include <algorithm>#include <queue>using namespace std;#define MAXN    1000+100#define MOD     1000000007int m, n;int dir[4][2] = {    {1,0},    {-1,0},    {0,1},    {0,-1}};char mp[MAXN][MAXN];int vis[MAXN][MAXN];int sz[MAXN][MAXN];struct node{    int x, y;    node() {};    node(int xx,int yy) :x(xx),y(yy){};}nnode;int bfs(int i,int j){    queue<node> q,visited;    vis[i][j] = 1;    int cnt = 1;    nnode.x = i; nnode.y = j;    q.push(nnode);    visited.push(nnode);    while (!q.empty())    {        nnode = q.front();        for (int i = 0; i < 4; i++)        {            int next_x = nnode.x + dir[i][0];            int next_y = nnode.y + dir[i][1];            if (next_x >= 0 && next_x < n&&next_y >= 0 && next_y < m&&mp[next_x][next_y] == '.'&&!vis[next_x][next_y])            {                vis[next_x][next_y] = 1;                cnt++;                q.push(node(next_x, next_y));                visited.push(node(next_x, next_y));            }        }        q.pop();    }    while (!visited.empty())    {        nnode = visited.front();        vis[nnode.x][nnode.y] = i*m+j+1;        sz[nnode.x][nnode.y] = cnt;        visited.pop();    }    return cnt;}int NotFind(long long arr[],int len,int key){    for (int i = 0; i < len; i++)        if (arr[i] == key)            return 0;    return 1;}int main(){    memset(sz, 0, sizeof(sz));    memset(vis, 0, sizeof(vis));    scanf("%d%d", &n, &m);    getchar();    for (int i = 0; i < n; i++)        scanf("%s",mp[i]);    for (int i = 0; i < n; i++)    {        for (int j = 0; j < m; j++)        {            if (mp[i][j] == '.'&&!vis[i][j])            {                bfs(i,j);            }        }    }    for (int i = 0; i < n; i++)    {        for (int j = 0; j < m; j++)        {            if (mp[i][j] == '*')            {                mp[i][j] = '0';                long long cnt = 1;                long long visited[5];                int viscnt = 0;                if (i - 1 >= 0 && NotFind(visited, viscnt, vis[i - 1][j])) { cnt += sz[i - 1][j]; visited[viscnt++] = vis[i - 1][j]; }                if (i + 1 < n && NotFind(visited, viscnt, vis[i + 1][j])) { cnt += sz[i + 1][j]; visited[viscnt++] = vis[i + 1][j]; }                if (j - 1 >= 0 && NotFind(visited, viscnt, vis[i][j - 1])) { cnt += sz[i][j - 1]; visited[viscnt++] = vis[i][j - 1]; }                if (j + 1 < m&& NotFind(visited, viscnt, vis[i][j + 1])) { cnt += sz[i][j + 1]; visited[viscnt++] = vis[i][j + 1]; }                cnt%= 10;                mp[i][j] += cnt;            }        }    }    for (int i = 0; i < n; i++)        printf("%s\n", mp[i]);    return 0;}

D.Longest k-Good Segment

这么水的题我竟然强行被坑了一天，，，，结果最后发现我只是宏定义的时候少加了个括号orz

解析

典型的two-pointers，先使用一个标记数组记录当前有哪些数，然后使用两个l,r标记当前左边界和右边界接着不断重复以下两步直到r==n-1:
1. 将右边界右移,直到当前不同的数的数目满足cnt>k
2. 将左边界右移，直到cnt<=k重新满足
期间维护一个满足条件的最大长度即可。

代码

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cassert>#include <algorithm>#include <queue>using namespace std;#define MAXN    500100#define MOD     1000000007int arr[MAXN];int vis[MAXN * 2];int main(){    memset(vis, 0, sizeof(vis));    int n, k;    scanf("%d%d",&n,&k);    for (int i = 0; i < n; i++)    {        scanf("%d",&arr[i]);    }    int maxlen = 0;    int ml=0, mr=0;    int cnt = 1;    int l = 0, r = 0;   //区间左端点和右端点下标    vis[arr[0]] = 1;    while (r<n)    {        int flag = (vis[arr[r + 1]] == 0) ? 1 : 0;        while (cnt + flag <= k)        {            cnt += flag;            vis[arr[r + 1]] ++;            r++;            if (r >= n - 1)                break;            flag = (vis[arr[r + 1]] == 0) ? 1 : 0;        }        if ((r - l + 1)>maxlen)        {            maxlen = (r - l + 1);            ml = l; mr = r;        }        vis[arr[l]]--;        if (vis[arr[l]] == 0)   //如果删除arr[l]后对应位置值为0，说明当前范围内不再有arr[l]，需要修改cnt            cnt--;         l++;    }    printf("%d %d\n",ml+1,mr+1);    return 0;}

声明

原文在CSDN上，链接http://blog.csdn.net/lincifer/article/details/50551482

还有我那连域名都还没有的博客上也有一份，链接http://115.28.240.133:8080/archives/96