51nod 1287 cannon (good exercise thinking)

Title: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1287

An array of positive integers of length m that represents the height of the terrain from left to right. Test a cannon, the shells parallel to the ground from left to right to fly, height is h, if the height of a terrain is greater than the height of the projectile flying H (A[i] >= h), the shells will be blocked and fall in I-1 place, then a[i-1] + 1. If H <= a[0], then this shell is invalid, if H > All A[i], this shell is not valid. Now the array B of the given n integers represents the height of the shells and calculates the final topography. For example: terrain height a = {1, 2, 0, 4, 3, 2, 1, 5, 7}, Shell height B = {2, 8, 0, 7, 6, 5, 3, 4, 5, 6, 5}, the resulting terrain height is: {2, 2, 2, 4, 3, 3, 5, 6, 7}. Input

Line 1th: 2 number m, n is separated by a space, respectively, the length of the array A and B (1 <= m, n <= 50000) 2nd to M + 1 lines: 1 numbers per line, indicating the corresponding terrain height (0 <= a[i] <= 1000000). The M + 2 to n + M + 1 lines, 1 numbers per line, indicate the height of the projectile (0 <= b[i] <= 1000000).

Output

Outputs a total of M rows, one number per line, corresponding to the final terrain height.

Input example

9 1112043215728076534565

Output example

222433567

Problem Solving Ideas:

The height of the shells is within 10^6, so we can find where it hit each height, and it's easy to think of the location of each height attack.

Must be non-diminishing, so it can be found in linear time, and then, for each projectile, it will only increase the height of the terrain in front of it by 1, which is

At most it is possible to change the position of a high attack, and it is better to judge the update.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int maxn=50000+1000;int a[maxn];int b[maxn];int c[maxn*20];int Main () {    int m,n;    scanf ("%d%d", &n,&m);    for (int i=0;i<n;i++)    scanf ("%d", &a[i]);    for (int j=0;j<m;j++)    scanf ("%d", &b[j]);    int h=0;    for (int i=0;i<m;i++)    H=max (H,b[i]);    int j=0;    for (int. i=0;i<=h;i++)    {        while (a[j]<i&& (J<n))        j + +;        c[i]=j;    }    for (int i=0;i<m;i++)    {        int te=c[b[i]];        if (te>0&&te<n)        {            a[te-1]++;            if (c[a[te-1]]>=te)            c[a[te-1]]=te-1;        }    }    for (int i=0;i<n;i++)    {        printf ("%d\n", A[i]);    }    return 0;}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

51nod 1287 cannon (good exercise thinking)