dy476813515的专栏

Pie Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 80   Accepted Submission(s) : 28Problem DescriptionMy birthday is coming up and tradit

Strange fuctionTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2949    Accepted Submission(s): 2177Problem DescriptionNow, here is a

Can you solve this equation?Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7689    Accepted Submission(s): 3574Problem DescriptionN

二分搜索--折半查找都是一个意思，

默默地超时了。。。我调了好久呢。。。不过还是有所有收获，1.二分可以用来查某数字x是否在递增序列中low=0;high=n-1;while(high>=low){    mid=(low+high)/2;    if(x>buf[mid]){        low=mid+1;    }    else if(x        high=mid-1;

//title：2141 Can you find it? //问题分类：二分，先将ab相加为s，判断 s[i]+c[j]?=x//即用二分查询找s中是否存在x-c[j] //时间复杂度：500(查找次数) *log2(250000)（2的20次为100万），约等于500*20 #include#include#includeusing namespace std;

//title：HDU2289 Cup//问题分类：二分解方程，和2199判断方法，1969思路很像，就是公式比较复杂 #include#includedouble pi = acos(-1.0);//圆台体积：v=pi*h(R*R+R*r+r*r)/3double getv(double r,double R,double H,double h){    double