Dilthey's Blog

二分查找。#includeint n,t,a[100000+5];bool judge(int num){ int sum=0; for(int j=1;j<=num;j++) sum+=a[j]; if(sum<=t) return true;//先把第一本书作为开始，读num本书，算出耗时 for(int i=2;i<=n-num+1;i++) { sum=sum-a[

题目大意：题目很长哈，简单来说，数轴上有n个石子，第i个石头的坐标为Di，现在要从0跳到L，每次条都从一个石子跳到相邻的下一个石子。现在FJ允许你移走M个石子，问移走这M个石子后，相邻两个石子距离的最小值的最大值是多少，因为移动有很多种方法，每一种有一个最小值距离，这些最小值里面的最大值（我之所以解释最后一句，是因为我兄弟他没懂--其实不是我翻译的，嘿嘿）。思路：在这里二分的对象当然是距离咯，

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5505    Accepted Submission(s): 1320Problem DescriptionToday is Yukari's n-th b

time limit per test5 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe main road in Bytecity is a straight line from south to north

time limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputLittle Nastya has a hobby, she likes to remove some letters from

DescriptionFarmer John has built a new long barn, with N (2 His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the

抽象来说就是求逆序对的问题，虽然放在树状数组的章节里，但不一定要用树状数组，归并排序也可以……#include#include#define MAX 1000*1000+3using namespace std;struct Road{ int u,v;}road[MAX],tmp[MAX];int n,m,k;long long cnt;bool cmp(Road a,Ro