zoukankan      html  css  js  c++  java
  • hdu 1034 (preprocess optimization, property of division to avoid if, decreasing order process) 分类: hdoj 2015-06-16 13:32 39人阅读 评论(0) 收藏

    IMO, version 1 better than version 2, version 2 better than version 3.
    make some preprocess to make you code simple and efficient. Here divide the input by 2, so you don’t have to do dividsion on each loop.
    version 1 is best
    thanks to
    http://www.cnblogs.com/kuangbin/archive/2012/06/03/2532690.html

    #include <cstdio>
    #include <algorithm>
    
    #define MAXSIZE 10000
    
    int candies[MAXSIZE];
    
    int main() {
        //freopen("input.txt","r",stdin);
        int n, cnt_whistle, i, tmp;
        while(scanf("%d",&n)!=EOF && n>0) {
            for(i=0;i<n;++i) { scanf("%d",&tmp); candies[i]=tmp>>1; }
            for(--n, cnt_whistle=1;;++cnt_whistle) {
                tmp=candies[n];
                for(i=n;i>0;--i) {
                    candies[i]=(candies[i-1]+candies[i]+1)>>1;
                }
                candies[0]=(tmp+candies[0]+1)>>1;
                for(i=1;i<=n && candies[i-1]==candies[i];++i) ;
                if(i-n==1) break;
            }
            printf("%d %d
    ",cnt_whistle,candies[0]<<1);
        }
        return 0;
    }

    use division’s floor property (5/2=2) to avoid if sentences. in version 2 here

    t2=candies[i];
    candies[i]=(t1+t2+1)>>1;
    t1=t2;

    if don’t divide input by 2, the code will be

    t2=candies[i]>>1;
    candies[i]=t1+t2;
    if(candies[i]%2) ++candies[i];
    t1=t2;

    how to avoid process the border conditions? version 2 better than version 3 on this aspect. version 1 accomplish this, thanks to
    http://www.acmerblog.com/hdu-1034-candy-sharing-game-1285.html

    version 2

    #include <cstdio>
    #include <algorithm>
    
    #define MAXSIZE 10000
    
    int candies[MAXSIZE];
    
    int main() {
        //freopen("input.txt","r",stdin);
        int n, cnt_whistle, i, t1,t2;
        while(scanf("%d",&n)!=EOF && n>0) {
            for(i=0;i<n;++i) { scanf("%d",&t1); candies[i]=t1>>1; }
            for(--n, cnt_whistle=1;;++cnt_whistle) {
                t1=candies[n];
                for(i=0;i<=n;++i) {
                    t2=candies[i];
                    candies[i]=(t1+candies[i]+1)>>1;
                    t1=t2;
                }
                for(i=1;i<=n && candies[i-1]==candies[i];++i) ;
                if(i-n==1) break;
            }
            printf("%d %d
    ",cnt_whistle,candies[0]<<1);
        }
        return 0;
    }
    

    version 3

    #include <cstdio>
    #include <algorithm>
    
    #define MAXSIZE 10000
    
    int candies[MAXSIZE];
    int cand_temp[MAXSIZE];
    
    int main() {
        //freopen("input.txt","r",stdin);
        int n, cnt_whistle, i, *p,*q;
        while(scanf("%d",&n)!=EOF && n>0) {
            for(i=0;i<n;++i) { scanf("%d",&t1); candies[i]=t1>>1; }
            p=candies, q=cand_temp;
            for(--n, cnt_whistle=1;;++cnt_whistle) {
                for(i=1;i<=n;++i) {
                    q[i]=(p[i-1]+p[i]+1)>>1;
                }
                q[0]=(p[n]+p[0]+1)>>1;
                std::swap(p,q);
                for(i=1;i<=n && p[i-1]==p[i];++i) ;
                if(i-n==1) break;
            }
            printf("%d %d
    ",cnt_whistle,p[0]<<1);
        }
        return 0;
    }
    

    p.s. when you use swap tricks (as in version 3), be careful with the check points, e.g. the initialization part, the conclusion part, especially the items are used by multiple times.
    here is mistake I made, 1st, initilization error
    position of

    cnt_whistle=1;

    mistakely put it to declaration part rather than just befor the for loop;
    2nd, the third parameter of printf
    mistakely wirite

    printf("%d %d
    ",cnt_whistle,candies[0]<<1);

    which should be

    printf("%d %d
    ",cnt_whistle,p[0]<<1);

    版权声明:本文为博主原创文章,未经博主允许不得转载。// p.s. If in any way improment can be achieved, better performance or whatever, it will be well-appreciated to let me know, thanks in advance.

  • 相关阅读:
    Linux 文件权限
    spak数据倾斜解决方案
    逻辑时钟
    kafka入门
    深入学习MySQL事务:ACID特性的实现原理
    程序员的诗
    java技术突破要点
    一个请求过来都经历了什么
    如何保持长时间高效学习
    你的系统如何支撑高并发
  • 原文地址:https://www.cnblogs.com/qeatzy/p/4716230.html
Copyright © 2011-2022 走看看