AIM Tech Round 3 (Div. 2) -- D. Recover the String (构造字符串)
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题目
题意:
给你一个只包含01的字符串,告诉你00,01,10,11子字符串的个数,还原这个字符串。不能存在则输出Impossible.
Examples
input
1 2 3 4
output
Impossible
input
1 2 2 1
output
0110
从00和11的个数,可以算出字符串中0和1的个数,然后即可构造。但当a00,a11中有为0的情况下,需要特殊考虑(代码水平问题….各种情况写的乱了,最后还是没有a….)。
#include <iostream>#include <cmath>using namespace std;long long a00, a01, a10, a11;int main() { bool flag = true; cin >> a00 >> a01 >> a10 >> a11; long long o = -1, z = -1; for(long long i = 1; i * (i - 1) / 2 <= (a00+1); i++){ if(i * (i - 1) == a00 * 2){ z = i; } } for(long long i = 1; i * (i - 1) / 2 <= (a11+1); i++){ if(i * (i - 1) == a11 * 2){ o = i; } } if(a00 == 0 && a01 == 0 && a10 == 0 && a11 == 0){ cout << '1' << endl; return 0; } if(a00 == 0){ if(a01 || a10) z = 1; else z = 0; } if(a11 == 0){ if(a10 || a01) o = 1; else o = 0; } if(o == -1 || z == -1) flag = false; if(a01 > z * o || a10 > z * o) flag = false; if(a01 + a10 != z * o) flag = false; if(flag) { long long t1 = 0, t2 = 0; while(o + z) { if(t1 + o > a01){ cout << '1'; a10 -= z; o--; } else{ cout << '0'; t1 += o; z--; } } } else { cout << "Impossible"; } cout << endl; return 0;}
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