CaoHaha's staff
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 514 Accepted Submission(s): 299
Problem Description "You shall not pass!"
After shouted out that,the Force Staff appered in CaoHaha's hand.
As we all know,the Force Staff is a staff with infinity power.If you can use it skillful,it may help you to do whatever you want.
But now,his new owner,CaoHaha,is a sorcerers apprentice.He can only use that staff to send things to other place.
Today,Dreamwyy come to CaoHaha.Requesting him send a toy to his new girl friend.It was so far that Dreamwyy can only resort to CaoHaha.
The first step to send something is draw a Magic array on a Magic place.The magic place looks like a coordinate system,and each time you can draw a segments either on cell sides or on cell diagonals.In additional,you need 1 minutes to draw a segments.
If you want to send something ,you need to draw a Magic array which is not smaller than the that.You can make it any deformation,so what really matters is the size of the object.
CaoHaha want to help dreamwyy but his time is valuable(to learn to be just like you),so he want to draw least segments.However,because of his bad math,he needs your help.
Input The first line contains one integer T(T<=300).The number of toys.
Then T lines each contains one intetger S.The size of the toy(N<=1e9).
Output Out put T integer in each line ,the least time CaoHaha can send the toy.
Sample Input
5
1
2
3
4
5
Sample Output
4
4
6
6
7
Source 2017中国大学生程序设计竞赛 - 网络选拔赛
Recommend liuyiding
只要能找到样例中的5个面积单位 7条边剩下的就好办了。5个斜边+2个正边组成的3个整正方形 + 5个三角形
推出来 边数 最大面积 4 2 6 4 7 5 8 8 9 10 10 12 11 14 12 18
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
using namespace std;
int shit[1000000];
int main() {
int b;
shit[4] = 2;
shit[6] = 4;
shit[7] = 5;
for (int i = 8; i <= 1000000; ++i) {
if (i % 4 == 0) {
b = i / 4;
shit[i] = b * b * 2;
} else if (i % 4 == 1) {
shit[i] = b - 1 + shit[i - 1];
} else if (i % 4 == 2) {
shit[i] = b * 2 + shit[i - 2];
} else {
shit[i] = b + shit[i - 1];
}
if (shit[i] > 1e9) {
break;
}
}
int t;
scanf("%d", &t);
while (t--) {
int x;
scanf("%d", &x);
for (int i = 4; i <= 1000000; ++i) {
if (shit[i] >= x) {
printf("%d\n", i);
break;
}
}
}
}