Hdu 5047 Sawtooth（数学+大整数）

题目链接

Sawtooth

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 920 Accepted Submission(s): 343



[align=left]Problem Description[/align]
Think about a plane:

● One straight line can divide a plane into two regions.

● Two lines can divide a plane into at most four regions.

● Three lines can divide a plane into at most seven regions.

● And so on...

Now we have some figure constructed with two parallel rays in the same direction, joined by two straight segments. It looks like a character “M”. You are given N such “M”s. What is the maximum number of regions that these “M”s can divide a plane ?



[align=left]Input[/align]
The first line of the input is T (1 ≤ T ≤ 100000), which stands for the number of test cases you need to solve.

Each case contains one single non-negative integer, indicating number of “M”s. (0 ≤ N ≤ 1012)

[align=left]Output[/align]
For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then an integer that is the maximum number of regions N the “M” figures can divide.

[align=left]Sample Input[/align]

2
1
2

[align=left]Sample Output[/align]

Case #1: 2
Case #2: 19

[align=left]Source[/align]
2014 ACM/ICPC Asia Regional Shanghai Online

题意：由两条平行的射线和两条线段构成的类似 ‘M’ 的图形，现在有n个，任意放，问最多能将平面分成多少个区域？

题解：由欧拉公式可知：对于一个几何结构的点数+面数-边数+1=空间维数

这题本来是个二维图，因为有射线，我们可以虚拟一个无穷远的点，把图形看成三维。那么问题就是求新图形的面数。

面数=边数+空间维数-点数-1；

n个‘M’相交，边数最多为：

一条线最多被切成：4*（n-1）+1 条线段

每个‘M’4条线段，n个‘M’的总线段就是：

16*n^2-12*n

空间维度为：3

点数最多为：

两个‘M’相交最多产生 16个点，一个‘M’自身有3个点，所有’M‘两两相交的话，再加上无穷远那个点，总点数是：

16*n*(n-1)/2+3*n+1=8*n^2-5*n+1

最后可得面数=8*n^2-7*n+1。

数据很大要用高精度，由于最后答案最大是10^24的数量级，也可以用两个long long 来拼成一个大整数。

高精度解法的代码：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<queue>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<math.h>
#include<stack>
typedef long long LL;
using namespace std;
struct BigInt
{
const static int mod = 10000;
const static int DLEN = 4;
int a[600],len;
BigInt()
{
memset(a,0,sizeof(a));
len = 1;
}
BigInt(long long v)
{
memset(a,0,sizeof(a));
len = 0;
do
{
a[len++] = v%mod;
v /= mod;
}while(v);
}
BigInt(const char s[])
{
memset(a,0,sizeof(a));
int L = strlen(s);
len = L/DLEN;
if(L%DLEN)len++;
int index = 0;
for(int i = L-1;i >= 0;i -= DLEN)
{
int t = 0;
int k = i - DLEN + 1;
if(k < 0)k = 0;
for(int j = k;j <= i;j++)
t = t*10 + s[j] - '0';
a[index++] = t;
}
}
BigInt operator +(const BigInt &b)const
{
BigInt res;
res.len = max(len,b.len);
for(int i = 0;i <= res.len;i++)
res.a[i] = 0;
for(int i = 0;i < res.len;i++)
{
res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);
res.a[i+1] += res.a[i]/mod;
res.a[i] %= mod;
}
if(res.a[res.len] > 0)res.len++;
return res;
}
BigInt operator *(const BigInt &b)const
{
BigInt res;
for(int i = 0; i < len;i++)
{
int up = 0;
for(int j = 0;j < b.len;j++)
{
int temp = a[i]*b.a[j] + res.a[i+j] + up;
res.a[i+j] = temp%mod;
up = temp/mod;
}
if(up != 0)
res.a[i + b.len] = up;
}
res.len = len + b.len;
while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;
return res;
}
void output()
{
printf("%d",a[len-1]);
for(int i = len-2;i >=0 ;i--)
printf("%04d",a[i]);
printf("\n");
}
}ans;
int main()
{
int t,cas=1;
LL n;
scanf("%d",&t);
while(t--)
{
scanf("%I64d",&n);
ans=BigInt(8*n-7)*BigInt(n)+BigInt(1);
printf("Case #%d: ",cas++);
ans.output();
}
return 0;
}

用两个long long 拼大整数的做法如下：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<queue>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<math.h>
#include<stack>
typedef long long LL;
using namespace std;
const int mod=1000000;
LL n;
int main()
{
int t,cas=1;
scanf("%d",&t);
while(t--)
{
scanf("%I64d",&n);
LL ltem,rtem;
ltem=(8*n-7)/mod,rtem=(8*n-7)%mod;
LL lans=0,rans;
rans=(rtem*n)%mod;
lans=ltem*n+(rtem*n)/mod;
rans=(rans+1)%mod;
lans=lans+(rans+1)/mod;
printf("Case #%d: ",cas++);
if(lans)
{
printf("%I64d%06I64d\n",lans,rans);
}
else
printf("%I64d\n",rans);
}
return 0;
}