1555: Inversion Sequence
Time Limit: 2 Sec Memory Limit:
256 MB
Submit: 360 Solved: 121
Description
For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j and greater to j at the same time. The sequence a1, a2, a3, … , aN is referred to as the inversion sequence of the original sequence (i1, i2, i3,
… , iN). For example, sequence 1, 2, 0, 1, 0 is the inversion sequence of sequence 3, 1, 5, 2, 4. Your task is to find a full permutation of 1~N that is an original sequence of a given inversion sequence. If there is no permutation meets the conditions please
output “No solution”.
Input
There are several test cases.
Each test case contains 1 positive integers N in the first line.(1 ≤ N ≤ 10000).
Followed in the next line is an inversion sequence a1, a2, a3, … , aN (0 ≤ aj < N)
The input will finish with the end of file.
Output
For each case, please output the permutation of 1~N in one line. If there is no permutation meets the conditions, please output “No solution”.
Sample Input
5 1 2 0 1 0 3 0 0 0 2 1 1
Sample Output
3 1 5 2 4 1 2 3 No solution
HINT
给你一个序列a[n],要你按照要求去构造一个序列b。
序列a[i]表示序列b中的i前面有a[i]个数比i大。
#include<cstdio> #include<iostream> #include<algorithm> #include<cmath> #include<vector> #define eps 1e-8 #define N 10100 using namespace std; int n; int a[N]; int main() { while(~scanf("%d",&n)) { for(int i=1; i<=n; i++) scanf("%d",&a[i]); vector<int>vec; bool flag=1; for(int i=n; i>=1; i--) { if(a[i]>vec.size()) { flag=0; break; } vec.insert(vec.begin()+a[i],i); } if(!flag)printf("No solution\n"); else { for(int i=0; i<vec.size(); i++) { if(i<vec.size()-1) printf("%d ",vec[i]); else printf("%d\n",vec[i]); } } } return 0; }