2013
11-27

# Simple Algorithmetics

One part of the new WAP portal is also a calculator computing expressions with very long numbers. To make the output look better, the result is formated the same way as is it usually used with manual calculations.
Your task is to write the core part of this calculator. Given two numbers and the requested operation, you are to compute the result and print it in the form specified below. With addition and subtraction, the numbers are written below each other. Multiplication is a little bit more complex: first of all, we make a partial result for every digit of one of the numbers, and then sum the results together.

There is a single positive integer T on the first line of input. It stands for the number of expressions to follow. Each expression consists of a single line containing a positive integer number, an operator (one of +, – and *) and the second positive integer number. Every number has at most 500 digits. There are no spaces on the line. If the operation is subtraction, the second number is always lower than the first one. No number will begin with zero.

For each expression, print two lines with two given numbers, the second number below the first one, last digits (representing unities) must be aligned in the same column. Put the operator right in front of the first digit of the second number. After the second number, there must be a horizontal line made of dashes (-).For each addition or subtraction, put the result right below the horizontal line, with last digit aligned to the last digit of both operands.

For each multiplication, multiply the first number by each digit of the second number. Put the partial results one below the other, starting with the product of the last digit of the second number. Each partial result should be aligned with the corresponding digit. That means the last digit of the partial product must be in the same column as the digit of the second number. No product may begin with any additional zeros. If a particular digit is zero, the product has exactly one digit — zero. If the second number has more than one digit, print another horizontal line under the partial results, and then print the sum of them.

There must be minimal number of spaces on the beginning of lines, with respect to other constraints. The horizontal line is always as long as necessary to reach the left and right end of both numbers (and operators) right below and above it. That means it begins in the same column where the leftmost digit or operator of that two lines (one below and one above) is. It ends in the column where is the rightmost digit of that two numbers. The line can be neither longer nor shorter than specified.

Print one blank line after each test case, including the last one.

4
12345+67890
324-111
325*4405
1234*4

 12345
+67890
------
80235

324
-111
----
213

325
*4405
-----
1625
0
1300
1300
-------
1431625

1234
*4
----
4936

#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;
const int MAX = 210;
struct line
{
double l,r,h;
int f;
line(){}
line(double a,double b,double c,int d):l(a),r(b),h(c),f(d){}
}a[MAX<<1];
double sum[MAX<<2],x0[MAX];
int cov[MAX<<2];
map<double,int>m;
int cmp(const line &a,const line &b)
{
return a.h<b.h;
}
void push_up(int l,int r,int rt)
{
if(cov[rt]) sum[rt] = x0[r+1] - x0[l];
else if(l==r) sum[rt]=0;
else sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void update(int c,int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
cov[rt]+=c;
push_up(l,r,rt);
return;
}
int mid = r+l>>1;
if(L<=mid) update(c,L,R,l,mid,rt<<1);
if(R>mid) update(c,L,R,mid+1,r,rt<<1|1);
push_up(l,r,rt);
}
int main()
{
int n;
double x1,x2,y1,y2;
int cas = 0;
while(scanf("%d",&n)&&n)
{
int cnt = 0;
int flag = 0;

for(int i=0; i<n; i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
a[cnt++] = line(x1,x2,y1,1);
a[cnt++] = line(x1,x2,y2,-1);
if(m[x1]==0) m[x1] = ++flag;
if(m[x2]==0) m[x2] = ++flag;
}
map<double,int>::iterator iter;
int k = 0;
for(iter=m.begin(); iter!=m.end(); iter++)
{
x0[++k] = iter->first;
iter->second = k;
}
sort(a,a+cnt,cmp);
double ans = 0;
memset(sum,0,sizeof(sum));
memset(cov,0,sizeof(cov));
for(int i=0; i<cnt-1; i++)
{
int l = m[a[i].l];
int r = m[a[i].r]-1;
update(a[i].f,l,r,1,flag,1);
ans+=(a[i+1].h-a[i].h)*sum[1];
}
printf("Test case #%d\n",++cas);
printf("Total explored area: %.2lf\n\n",ans);
m.clear();
}
return 0;
}

1. 有两个重复的话结果是正确的，但解法不够严谨，后面重复的覆盖掉前面的，由于题目数据限制也比较严，所以能提交通过。已更新算法

2. 如果两个序列的最后字符不匹配（即X [M-1]！= Y [N-1]）
L（X [0 .. M-1]，Y [0 .. N-1]）= MAX（L（X [0 .. M-2]，Y [0 .. N-1]），L（X [0 .. M-1]，Y [0 .. N-1]）
这里写错了吧。

3. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
O(n) = O(n-1) + O(n-2) + ….
O(n-1) = O(n-2) + O(n-3)+ …
O(n) – O(n-1) = O(n-1)
O(n) = 2O(n-1)

4. 老实说，这种方法就是穷举，复杂度是2^n，之所以能够AC是应为题目的测试数据有问题，要么数据量很小，要么能够得到k == t，否则即使n = 30，也要很久才能得出结果，本人亲测