zoukankan      html  css  js  c++  java
  • hdoj--1556--Color the ball(模拟&&树状数组)

    Color the ball

    Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 13727    Accepted Submission(s): 6903



    Problem Description
    N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了,你能帮他算出每个气球被涂过几次颜色吗?
     

    Input
    每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行,每行包括2个整数a b(1 <= a <= b <= N)。
    当N = 0,输入结束。
     

    Output
    每个测试实例输出一行,包括N个整数,第I个数代表第I个气球总共被涂色的次数。
     

    Sample Input
    3 1 1 2 2 3 3 3 1 1 1 2 1 3 0
     

    Sample Output
    1 1 1 3 2 1
     

    Author
    8600
     

    Source
     

    Recommend
    LL   |   We have carefully selected several similar problems for you:  1542 1394 1698 1255 2795

    先来一组大神的代码,真神奇的模拟,思路太好了

    #include<stdio.h>
    #include<string.h>
    int num[1000010];
    int main()
    {
    	int n;
    	while(scanf("%d",&n),n)
    	{
    		memset(num,0,sizeof(num));
    		int m=0;
    		for(int i=0;i<n;i++)
    		{
    			int a,b;
    			scanf("%d%d",&a,&b);
    			num[a]++;
    			num[b+1]--;
    		}
    		for(int i=1;i<n;i++)
    		{
    			m+=num[i];
    			printf("%d ",m);
    		}
    		printf("%d
    ",m+num[n]);
    	}
    	return 0;
    }

    树状数组
    #include<stdio.h>
    #include<string.h>
    int num[100010];
    int n;
    void add(int x,int k)
    {
    	while(x>=1)
    	{
    		num[x]+=k;
    		x-=x&(-x);
    	}
    }
    int sum(int x)
    {
    	int s=0;
    	while(x<=n)
    	{
    		s+=num[x];
    		x+=x&(-x);
    	}
    	return s;
    }
    int main()
    {
    	while(scanf("%d",&n),n)
    	{
    		memset(num,0,sizeof(num));
    		for(int i=0;i<n;i++)
    		{
    			int a,b;
    			scanf("%d%d",&a,&b);
    			add(b,1);
    			add(a-1,-1);
    		}
    		for(int i=1;i<=n;i++)
    		{
    			if(i>1)
    			printf(" ");
    			printf("%d",sum(i));
    		}
    		printf("
    ");
    	}
    	return 0;
    }

     
  • 相关阅读:
    基于Python的人脸动漫转换
    let 与 var的区别
    【LeetCode】汇总
    【HDU】4632 Palindrome subsequence(回文子串的个数)
    【算法】均匀的生成圆内的随机点
    【LeetCode】725. Split Linked List in Parts
    【LeetCode】445. Add Two Numbers II
    【LeetCode】437. Path Sum III
    【LeetCode】222. Count Complete Tree Nodes
    【LeetCode】124. Binary Tree Maximum Path Sum
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273671.html
Copyright © 2011-2022 走看看