Codeforces Round #318

A. Bear and Elections
每次通过优先队列选取最大的数值，并与第一个人的票数l进行比较，大于l，自减1，并记录大于1的次数即为需要更改票数的人数。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <cmath>#include <algorithm>#include <set>#include <map>#include <queue>#include <ctime>#define pb push_back#define ll long long#define mp make_pair#define f first#define s second#define pii pair < int, int >#define ull unsigned long long#define pll pair < ll, ll >#define forit(s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)#define all(s) s.begin(), s.end()const int inf = (1ll << 31) - 1;const int maxn = (int) 1e5 + 10;#define max(a, b) ((a) > (b) ? (a) : (b))#define min(a, b) ((a) < (b) ? (a) : (b)) using namespace std;int n = 0, l = 0;int ans = 0;int main(){    //freopen("data_A.txt", "r", stdin);    priority_queue<int> s;    ans = 0;    scanf("%d%d", &n, &l);    for (int i = 1; i < n; i++)    {        int a = 0;        scanf("%d", &a);        s.push(a);    }    while(s.top() >= l)    {        l++;        int a = 0;        a = s.top() - 1;        s.pop();        s.push(a);        ans++;    }    printf("%d\n", ans);    return 0; }

再或者手写优先队列：每次找出数组中的最大值：

#include<iostream>#include<fstream>using namespace std;int n = 0, a[110];int rec = 0, ans = 0;int main(){    //freopen("data_A.txt", "r", stdin);    while(scanf("%d", &n) != EOF)    {        ans = 0;        for (int i = 0; i < n; i++)        {            scanf("%d", &a[i]);        }        while(true)        {            rec = 0;            for (int i = 1; i < n; i++)            {                if (a[i] >= a[rec])                {                    rec = i;                }            }            if (rec == 0) break;            a[0]++;            a[rec]--;            ans++;        }        printf("%d\n", ans);    }    return 0;}

B. Bear and Three Musketeers
n个人里面选三个人，除了这三个人之外， 求其他的有多少个人认识这三个人。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <cmath>#include <algorithm>#include <set>#include <map>#include <queue>#include <ctime>#define pb push_back#define ll long long#define mp make_pair#define f first#define s second#define pii pair < int, int >#define ull unsigned long long#define pll pair < ll, ll >#define forit(s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)#define all(s) s.begin(), s.end()const int inf = (1ll << 31) - 1;const int maxn = (int) 1e5 + 10;#define max(a, b) ((a) > (b) ? (a) : (b))#define min(a, b) ((a) < (b) ? (a) : (b)) using namespace std;int n = 0, m = 0, matrix[4100][4100], indegree[4100]; int main(){    //freopen("data_B.txt", "r", stdin);    while(scanf("%d%d", &n, &m) != EOF)    {        memset(matrix, 0, sizeof(matrix));        memset(indegree, 0, sizeof(indegree));        for (int i= 0; i < m; i++)        {            int a = 0, b = 0;            scanf("%d%d", &a, &b);            matrix[a][b] = 1;            matrix[b][a] = 1;            indegree[a]++;            indegree[b]++;        }        int ans = inf;        for (int i = 1; i <= n; i++)        {            for (int j = i + 1; j <= n; j++)            {                if (matrix[i][j])                {                    for (int k = j + 1; k <= n; k++)                    {                        if (matrix[i][k] && matrix[j][k])                        {                            ans = min (ans, indegree[i] + indegree[j] + indegree[k]);                        }                    }                }            }        }         if (ans == inf)        {            printf("-1\n");        }else        {            printf("%d\n", ans - 6);        }    }    return 0; }

C. Bear and Poker
递归实现超时，只有改用其他方法。改成非递归的方式。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <cmath>#include <algorithm>#include <set>#include <map>#include <queue>#include <ctime>#define pb push_back#define ll long long#define mp make_pair#define f first#define s second#define pii pair < int, int >#define ull unsigned long long#define pll pair < ll, ll >#define forit(s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)#define all(s) s.begin(), s.end()const int inf = (1ll << 31) - 1;const int maxn = (int) 1e5 + 10;#define max(a, b) ((a) > (b) ? (a) : (b))#define min(a, b) ((a) < (b) ? (a) : (b)) using namespace std;int n = 0, a[110000]; int main(){    //freopen("data_C.txt", "r", stdin);    while(scanf("%d", &n) != EOF)    {        for (int i = 0; i < n; i++)        {            scanf("%d", &a[i]);        }        for (int i = 0; i < n; i++)        {            while (a[i] % 2 == 0)            {                a[i] /= 2;            }            while (a[i] % 3 == 0)            {                a[i] /= 3;            }        }        for (int i = 1; i < n; i++)        {            if (a[i] != a[0])            {                printf("No\n");                return 0;            }         }        printf("Yes\n");    }     return 0; }

D. Bear and Blocks
动态规划：从左到右以及从右到左进行遍历，找出其中的最小值。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <cmath>#include <algorithm>#include <set>#include <map>#include <queue>#include <ctime>#define pb push_back#define ll long long#define mp make_pair#define f first#define s second#define pii pair < int, int >#define ull unsigned long long#define pll pair < ll, ll >#define forit(s) for(__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)#define all(s) s.begin(), s.end()const int inf = (1ll << 31) - 1;const int maxn = (int) 1e5 + 10;#define max(a, b) ((a) > (b) ? (a) : (b))#define min(a, b) ((a) < (b) ? (a) : (b)) using namespace std;int l[110000], r[110000], dp[110000], ans = 0;int n = 0;int d[110000];int main(){    //freopen("data_D.txt", "r", stdin);    while(scanf("%d", &n) != EOF)    {        ans = 0;        for (int i = 0; i < n; i++)        {            scanf("%d", &d[i]);        }        l[0] = 1;        r[n - 1] = 1;        for (int i = 1; i < n; i++)        {            l[i] = min (l[i - 1] + 1, d[i]);        }        for (int i = n - 2; i > -1; i--)        {            r[i] = min (r[i + 1] + 1, d[i]);        }        for (int i = 0; i < n; i++)        {            dp[i] = min (l[i], r[i]);        }        for (int i = 0; i < n; i++)        {            ans = max (ans, dp[i]);        }        printf("%d\n", ans);    }    return 0; }