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ZOJ 3365 Integer Numbers
Integer Numbers
64-bit integer IO format: %lld Java class name: Main
The boy likes numbers. He has a sheet of paper. He have written a sequence of consecutive integer numbers on the sheet. The boy likes them.
But then the girl came. The girl is cruel. She changed some of the numbers.
The boy is disappointed. He cries. He does not like all these random numbers. He likes consecutive numbers. He really likes them. But his numbers are not consecutive any more. The boy is disappointed. He cries.
Help the boy. He can change some numbers. He would not like to change many of them. He would like to change as few as possible. He cannot change their order. He would like the numbers to be consecutive again. Help the boy.
Input
The first line of the input file contains n --- the number of numbers in the sequence (1 ≤ n ≤ 50000). The next line contains the sequence itself --- integer numbers not exceeding 109 by their absolute values.
There are multiple cases. Process to the end of file.
Output
Output the minimal number of numbers that the boy must change. After that output the sequence after the change.
Sample Input
65 4 5 2 1 8
Sample Output
33 4 5 6 7 8
Source
Author
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 51000;18 vector<int>g[maxn];19 int lisan[maxn],d[maxn],n;20 int main() {21 while(~scanf("%d",&n)){22 for(int i = 0; i < n; ++i){23 g[i].clear();24 scanf("%d",d+i);25 lisan[i] = d[i] - i;26 }27 sort(lisan,lisan+n);28 int cnt = 1;29 for(int i = 1; i < n; ++i)30 if(lisan[cnt-1] != lisan[i]) lisan[cnt++] = lisan[i];31 for(int i = 0; i < n; ++i){32 int index = lower_bound(lisan,lisan+cnt,d[i]-i)-lisan;33 g[index].push_back(i);34 }35 int idx = -1,mx = 0;36 for(int i = 0; i < cnt; ++i)37 if(g[i].size() > mx) mx = g[idx = i].size();38 printf("%d\n",n - g[idx].size());39 int tmp = g[idx][0];40 tmp = d[tmp] - tmp;41 for(int i = 0; i < n; ++i)42 printf("%d%c",tmp + i,i+1 == n?‘\n‘:‘ ‘);43 }44 return 0;45 }
ZOJ 3365 Integer Numbers