Many problems in Computer Science involve maximizing some measureaccording to constraints.

Consider a history exam in which students are asked to put severalhistorical events into chronological order. Students who order all theevents correctly will receive full credit, but how should partial creditbe awarded to students who incorrectly rank one or more of thehistorical events?

Some possibilities for partial credit include:

1. 1 point for each event whose rank matches its correct rank
2. 1 point for each event in the longest (not necessarily contiguous)sequence of events which are in the correct order relative to each other.

For example, if four events are correctly ordered 1 2 3 4 then theorder 1 3 2 4 would receive a score of 2 using the first method (events1 and 4 are correctly ranked) and a score of 3 using the second method(event sequences 1 2 4 and 1 3 4 are both in the correct order relativeto each other).

In this problem you are asked to write a program to score such questionsusing the second method.

Given the correct chronological order of n events as where denotes the ranking ofevent i in the correct chronological order and a sequence of studentresponses where denotes thechronological rank given by the student to event i; determine the length of the longest (not necessarily contiguous) sequenceof events in the student responses that are in the correct chronologicalorder relative to each other.

The first line of the input will consist of one integer n indicatingthe number of events with . The second line willcontain n integers, indicating the correct chronological order of nevents. The remaining lines will each consist of n integers with eachline representing a student's chronological ordering of the n events.All lines will contain n numbers in the range , with each numberappearing exactly once per line, and with each number separated fromother numbers on the same line by one or more spaces.

For each student ranking of events your program should print the scorefor that ranking. There should be one line of output for each studentranking.

`44 2 3 11 3 2 43 2 1 42 3 4 1`
`123`
`103 1 2 4 9 5 10 6 8 71 2 3 4 5 6 7 8 9 104 7 2 3 10 6 9 1 5 83 1 2 4 9 5 10 6 8 72 10 1 3 8 4 9 5 7 6`
`65109`

```#include

#include

using namespace std;int a[101],b[101],dp[101][101];int main(){
int t,n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>t;
a[i]=t;
}
for(int j=1;j<=n;j++)
{
cin>>t;
b[j]=t;
}
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(a[i]==b[j])
{
dp[i][j]=dp[i-1][j-1]+1;
}
else
{
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
}
cout<

<

return 0;}

```

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[***.249.68.18]2022年05月26日 10时26分27秒

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