CodeForces 708B Recover the String
CodeForces 708B Recover the String
B. Recover the String
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
For each string s consisting of characters '0' and '1' one can define four integers a00, a01, a10 and a11, where axy is the number ofsubsequences of length 2 of the string s equal to the sequence {x, y}.
In these problem you are given four integers a00, a01, a10, a11 and have to find any non-empty string s that matches them, or determine that there is no such string. One can prove that if at least one answer exists, there exists an answer of length no more than1 000 000.
Input
The only line of the input contains four non-negative integers a00, a01, a10 and a11. Each of them doesn't exceed 109.
Output
If there exists a non-empty string that matches four integers from the input, print it in the only line of the output. Otherwise, print "Impossible". The length of your answer must not exceed 1 000 000.
Examples
input
1 2 3 4output
Impossibleinput
1 2 2 1output
0110
可以根据a00和a11分别确定0和1的个数
注意边界情况：a00是0，0的个数可以为1，要根据a01的值判断
在判断impossible方面，首先a00和a11必须是n*(n+1)/2的形式，
其次a01+a10=n0*n1
然后构造一个符合条件的字符串
先把所有1放前面，所有0放后面，根据a01的个数将0向前移动
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>
using namespace std;
int a00,a01,a10,a11;
int n0,n1;
int main()
{
scanf("%d%d%d%d",&a00,&a01,&a10,&a11);
int b1=0,b2=0;
n0=1;n1=1;
if(a00==0&&a01==0&&a10==0&&a11==0)
{
printf("0\n");
return 0;
}
while(b1<a00) {b1+=n0;n0++;}
while(b2<a11) {b2+=n1;n1++;}
if(a00==0&&a01==0&&a10==0) n0=0;
if(a11==0&&a01==0&&a10==0) n1=0;
if(b1!=a00||b2!=a11||(a01+a10!=n1*n0))
{
printf("Impossible\n");
return 0;
}
if(a00==0&&a10==0&&a01==0)
{
for(int i=0;i<n1;i++)
printf("1");
printf("\n");
return 0;
}
if(a11==0&&a10==0&&a01==0)
{
for(int i=0;i<n0;i++)
printf("0");
printf("\n");
return 0;
}
int num1=a01/n1;
int num2=a01%n1;
for(int i=0;i<num1;i++)
printf("0");
for(int i=0;i<n1-num2;i++)
printf("1");
if(num1!=n0)
printf("0");
for(int i=0;i<num2;i++)
printf("1");
for(int i=0;i<n0-num1-1;i++)
printf("0");
printf("\n");
return 0;
}