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A Padovan string P(n) for a natural number n is defined as: P(0) = ‘X’ P(1) = ‘Y’ P(2) = ‘Z’ P(n) = P(n-2) + P(n-3), n>2 where + denotes string concatenation. For a string of the characters ‘X’ , ‘Y’ and ‘Z’ only, and given value of n, write a program that counts the number of occurrences of the string in the n-th Padovan string P(n). An example is given below. For n = 6 and the string ZY, the program should count the occurrences of ZY in P(6). P(0) = ‘X’ P(1) = ‘Y’ P(2) = ‘Z’ P(n) = P(n-2) + P(n-3), n>2 P(3) = P(1)+P(0) P(3) = YX P(4) = P(2)+P(1) P(4) = ZY P(5) = P(3)+P(2) P(5) = YXZ P(6) = P(4)+P(3) P(6) = ZYYX So, the number of occurrences of the string ZY in P(6) is 1.
#include stdio.h> #include conio.h> #include string.h> void main() { char a[50],b[25][25],l[25],temp1; int i=0,j=0,k=0,l1[25],c=0,c1=0,t=
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#include stdio.h> #include conio.h> #include string.h> void main() { int i=0,j=0,k=0,l=0; int a[3][3],temp[3][3]; clrscr();
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