hzau 1209 Deadline（贪心）

K．Deadline There are N bugs to be repaired and some engineers whose abilities are roughly equal. And an engineer can repair a bug per day. Each bug has a deadline A[i].

Question: How many engineers can repair all bugs before those deadlines at least? 1<=n<= 1e6. 1<=a[i] <=1e9

Input Description There are multiply test cases. In each case, the first line is an integer N , indicates the number of bugs. The next line is n integers indicates the deadlines of those bugs.

Output Description There are one number indicates the answer to the question in a line for each case.

Input 4 1 2 3 4

Output 1

排序，大于N的不用考虑，否则超时

贪心

 #include <bits/stdc++.h>
 using namespace std;

 const int MAXN = 1e6 + ;

 int a[MAXN];
 int b[MAXN];
 int tot;

 int main()
 {
     int n;
     int i;
     int j;
     int maxB;
     int lMaxB;

     while (~scanf("%d", &n)) {
         int n2 = n;
         for (i = ; i < n; ++i) {
             scanf("%d", &a[i]);
             if (a[i] >= n2) {
                 --i;
                 --n;
             }
         }
         //cout << n << endl;
         sort(a, a + n);

         tot = ;
         b[tot++] = ;
         for (i = ; i < n; ++i) {
             maxB = -;
             for (j = ; j < tot; ++j) {
                 if (b[j] < a[i] && b[j] > maxB) {
                     maxB = b[j];
                     lMaxB = j;
                     break;
                 }
             }

             if (maxB == -) {
                 b[tot++] = ;
             } else {
                 ++b[lMaxB];
             }

         }

         printf("%d\n", tot);
     }

     return ;
 }

但是为什么用set写超时呢，不是应该比数组遍历查找要快吗？

超时代码：

 #include <bits/stdc++.h>
 using namespace std;

 const int MAXN = 1e6 + ;

 int a[MAXN];

 struct cmp {
     bool operator()(int a, int b)
     {
         return a > b;
     }
 };
 multiset<int, cmp> st;

 int main()
 {
     int n;
     int i;
     multiset<int, cmp>::iterator it;
     int tmp;

     while (~scanf("%d", &n)) {
         st.clear();
         int n2 = n;
         for (i = ; i < n; ++i) {
             scanf("%d", &a[i]);
             if (a[i] >= n2) {
                 --i;
                 --n;
             }
         }
         //cout << n << endl;
         sort(a, a + n);

         st.insert();
         for (i = ; i < n; ++i) {
             it = st.upper_bound(a[i]);
             if (it == st.end()) {
                 st.insert();
             } else {
                 tmp = *it + ;
                 st.erase(it);
                 st.insert(tmp);
             }
         }

         printf("%d\n", st.size());

     }

     return ;
 }

其实这题可以用 任务数 / 天数，向上取整得到最少需要的人数，取最大值

 #include <bits/stdc++.h>
 using namespace std;

 const int MAXN = 1e6 + ;

 int maxA;
 int b[MAXN];//b[i]保存当天要完成的任务数
 int sum[MAXN];//sum[i]代表之前一共要完成任务数

 int main()
 {
     int n;
     int i;
     int a;
     int ans;

     while (~scanf("%d", &n)) {

         memset(b, , sizeof(b));
         for (i = ; i < n; ++i) {
             scanf("%d", &a);
             if (a > n) {
                 continue;
             }
             ++b[a];
             if (a > maxA) {
                 maxA = a;
             }
         }
         //cout << n << endl;

         sum[] = ;
         ans = ;//最小为1个工人...//初始化为0不对。因为有可能所有日期都大于n，那么结果应该为1，而不是0
         for (i = ; i <= maxA; ++i) {
             sum[i] = sum[i - ] + b[i];
             ans = max(ans, (sum[i] + i - ) / i);
         }

         printf("%d\n", ans);
     }

     return ;
 }