The 2016 ACM-ICPC Asia China-Final C (15/600)

Mr. Panda likes ice cream very much especially the ice cream tower. An ice cream tower consists
of K ice cream balls stacking up as a tower. In order to make the tower stable, the lower ice cream
ball should be at least twice as large as the ball right above it. In other words, if the sizes of the
ice cream balls from top to bottom are A0, A1, A2, · · · , AK−1, then A0 × 2 ≤ A1, A1 × 2 ≤ A2,
etc.
One day Mr. Panda was walking along the street and found a shop selling ice cream balls. There
are N ice cream balls on sell and the sizes are B0, B1, B2, · · · , BN−1. Mr. Panda was wondering
the maximal number of ice cream towers could be made by these balls.
Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test
case starts with a line consisting of 2 integers, N the number of ice cream balls in shop and K
the number of balls needed to form an ice cream tower. The next line consists of N integers
representing the size of ice cream balls in shop.
Output
For each test case, output one line containing “Case #x: y”, where x is the test case number
(starting from 1) and y is the maximal number of ice cream towers could be made.
Limits
• 1 ≤ T ≤ 100.
• 1 ≤ N ≤ 3 × 105
.
• 1 ≤ K ≤ 64.
• 1 ≤ Bi ≤ 1018
.
Sample input and output
Sample Input Sample Output
3
4 2
1 2 3 4
6 3
1 1 2 2 4 4
6 3
1 1 2 2 3 4
Case #1: 2
Case #2: 2
Case #3: 1

就是个普通的二分….枚举答案
看行不行
检测的时候只要一层一层的贪心就可以了。。更大的不行那小的也不行

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;long long n, m, tu[300001],bj[300001];long long jiance(long long k){    memset(bj, 0, sizeof(bj));    long long p = 0;    for (long long a = 1; a <= k; a++)bj[a] = tu[++p];    for (long long a = 1; a < m; a++)    {        for (long long b = 1; b <= k; b++)        {            p++;            while (p <= n&&tu[p] < 2 * bj[b])p++;            if (p > n)return 0;            bj[b] = tu[p];        }    }    return 1;}int main(){    long long T;    cin >> T;    long long u = 0;    while(T--)    {        cin >> n >> m;        for (long long a = 1; a <= n; a++)scanf("%lld", &tu[a]);        sort(tu + 1, tu + n + 1);        long long y = n / m, z = 0;        long long jieg = 0;        while (z <= y)        {            long long mid = (z + y) / 2;            if (jiance(mid))            {                jieg = mid;                z = mid + 1;            }            else y = mid - 1;        }        printf("Case #%lld: %lld\n", ++u, jieg);    }}