zoukankan      html  css  js  c++  java
  • 二分C

    <span style="color:#330099;">/*
    C - 二分 基础
    Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
    Submit
     
    Status
    Description
    It is very hard to wash and especially to dry clothes in winter. But Jane is a very smart girl. She is not afraid of this boring process. Jane has decided to use a radiator to make drying faster. But the radiator is small, so it can hold only one thing at a time.
    
    Jane wants to perform drying in the minimal possible time. She asked you to write a program that will calculate the minimal time for a given set of clothes.
    
    There are n clothes Jane has just washed. Each of them took ai water during washing. Every minute the amount of water contained in each thing decreases by one (of course, only if the thing is not completely dry yet). When amount of water contained becomes zero the cloth becomes dry and is ready to be packed.
    
    Every minute Jane can select one thing to dry on the radiator. The radiator is very hot, so the amount of water in this thing decreases by k this minute (but not less than zero — if the thing contains less than k water, the resulting amount of water will be zero).
    
    The task is to minimize the total time of drying by means of using the radiator effectively. The drying process ends when all the clothes are dry.
    
    Input
    The first line contains a single integer n (1 ≤ n ≤ 100 000). The second line contains ai separated by spaces (1 ≤ ai ≤ 109). The third line contains k (1 ≤ k ≤ 109).
    
    Output
    Output a single integer — the minimal possible number of minutes required to dry all clothes.
    
    Sample Input
    sample input #1 3
    2 3 9
    5 sample input #2 3
    2 3 6
    5
    Sample Output
    sample output #1 3 sample output #2 2
    2014.7.15
    */
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    #include <math.h>
    int water[100005], cloth, k;
    bool check(int m) {
        int cnt = 0; 
        for(int i = 0; i < cloth; i++) {
           if(water[i] <= m)   continue;
            else {
                cnt += (int)ceil((double)(water[i] - m) / (k-1));//
                if(cnt > m) return false;
            }
        }
        return true;
    }
       int main() {
        int maxi = 0;
        scanf("%d", &cloth);
        for(int i = 0; i < cloth; i++) {
            scanf("%d", water+i);
            if(water[i] > maxi) maxi = water[i];
        }
        scanf("%d", &k);
        if(k > 1) {
            int l = 1, r = maxi, mid;
            while(l <= r) {
                mid = (l+r) >> 1;
    
                if(check(mid))  r = mid - 1;
                else            l = mid + 1;
            }
            printf("%d
    ", l);
        }
        else    printf("%d
    ", maxi);
       return 0;
    }
    
    </span>

  • 相关阅读:
    递归
    lecture-11
    最近公共祖先LCA
    微软面试100题
    0-1背包问题
    ubuntu网络已禁用解决办法
    cors(cross-resource-oragin-sharing 跨域资源共享) 解决跨域问题
    本地修改域名对应的IP地址
    oracle 解锁用户被锁住
    oracle 改变表中 某列的数据类型(该列已有数据)
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254515.html
Copyright © 2011-2022 走看看