Taming the Herd

题目

• Problem Description
  Early in the morning, Farmer John woke up to the sound of splintering wood. It was the cows, and they were breaking out of the barn again!

  Farmer John was sick and tired of the cows’ morning breakouts, and he decided enough was enough: it was time to get tough. He nailed to the barn wall a counter tracking the number of days since the last breakout. So if a breakout occurred in the morning, the counter would be 0 that day; if the most recent breakout was 3 days ago, the counter would read 3. Farmer John meticulously logged the counter every day.

  The end of the year has come, and Farmer John is ready to do some accounting. The cows will pay, he says! But lo and behold, some entries of his log are missing!

  Farmer John is confident that the he started his log on the day of a breakout. Please help him determine, out of all sequences of events consistent with the log entries that remain, the minimum and maximum number of breakouts that may have take place over the course of the logged time.

• Input
  The first line contains a single integer N (1≤N≤100), denoting the number of days since Farmer John started logging the cow breakout counter.

  The second line contains N space-separated integers. The i-th integer is either −1, indicating that the log entry for day i is missing, or a non-negative integer a_i (at most 100), indicating that on day i the counter was at a_i.

• Output
  If there is no sequence of events consistent with Farmer John’s partial log and his knowledge that the cows definitely broke out of the barn on the morning of day 1, output a single integer −1. Otherwise, output two space-separated integers mm followed by M, where mm is the minimum number of breakouts of any consistent sequence of events, and M is the maximum.

• input
  4
  -1 -1 -1 1

• output
  2 3

题意

现有一个记录仪，每当有牛跑出来的時候则会记录0，如果当天没有牛跑出来则数字+1，第一天数字一定为0
不过有些数字丢失了，求出牛最小和最大跑出来的次数
如果有记录冲突，则输出-1

思路

从记录的最后往前遍历，将可以确定的数字，即>0的数据前面的数字更新，剩下的-1即为不确定的数字，不确定数字全部为0则为最大次数，全部不为0则为最小次数

代码

#include <bits/stdc++.h>
using namespace std;

int b[105];

int main() {
    int N;
    scanf("%d",&N);
    for(int n=1; n<=N; ++n)
        scanf("%d",b+n);

    int sum=0;
    for(int n=N; n>0; --n) {//更新可以确定的数据
        if( b[n]!=-1 ) {
            ++sum;
            if(b[n]!=0) {
                int i;
                for(i=n-1; i>=0 && b[i+1]!=0; --i) {
                    if(i==0) {
                        printf("-1\n");
                        return 0;
                    }
                    else if( b[i] == -1 or b[i] == b[i+1]-1) {
                        b[i]=b[i+1]-1;
                    }
                    else {
                        printf("-1\n");
                        return 0;
                    }
                }
                n=i+1;
            }
        }
    }

    if(b[1]>0) 
        printf("-1\n");
    else {
        if(b[1]==-1) ++sum,b[1]=0;
        int max=sum;
        for(int n=1; n<=N; ++n) {
            if(b[n]==-1)
                ++max;
        }
        printf("%d %d\n",sum,max);
    }
    return 0;
}