ACM practice every day (monkey peach problem)

Description

There are a bunch of peaches I don't know how to count. The monkeys eat half of them on the first day and eat one more. The next day, this method is used to eat more than half of the peaches. Every day, on the morning of the M day, the monkey found only one peach left. How many peaches were there? (M <29)

Input

The first row has an integer N, indicating that there are N groups of test data (starting from the second row, the data in each row is: day m );

Output

Each row of data is the total number of peaches.

Sample Input

2311

Sample output

226142

Common Solution
Analysis: The recurrence formula for the relationship between the total number of peaches in the previous day and the total number of peaches in the next day is a [M-1] = (a [m] + 1) x 2; we can find the total number of peaches on the first day by repeating m times.

#include <stdio.h>main(){ int i,n,m,sum; scanf("%d",&n); while(n--) {  sum=1;  scanf("%d",&m);  for (i=m;i>0;i--)   sum=2*(sum+1);  printf("%d\n",sum); }}

The recursive method changes the previous for loop to recursive call.

#include <stdio.h>int sum(int n){ if (n==0) return 1; else return 2*(sum(n-1)+1);}main(){ int m,n; scanf("%d",&m); while(m--) {  scanf("%d",&n);  printf("%d\n",sum(n)); }}

Simplest solutionAnalysis A [0] = 1a [1] = (a [0] + 1) * 2... A [n-1] = (a [N-2] + 1) * 2a [N] = (a [n-1] + 1) * 2 from where, the general formula for this series can be obtained: A [n] = 3 * (POW (2, n)-2
Then the program can be improved

# Include <stdio. h> main () {int M, N; scanf ("% d", & M); While (M --) {scanf ("% d", & N ); // 3 shifts n places to the left, which is equivalent to 3 * POW (2, n) printf ("% d \ n", (3 <n)-2 );}}

Description

Give you a disordered string containing lowercase letters (A -- Z) and some special characters. Please find out the number of all lowercase letters in the given string, take this number pair for 26 and output the number after the remainder in lowercase letters (0 for Z, 1 for A, 2 for B .... 25 corresponds to Y ).

Input

The first line is an integer N (1 <n <1000) indicating that the string M (1 <m <200) with N rows to be input

Output

Output the corresponding lowercase letters. Each lowercase letter occupies a single row.

Sample Input

2asdasl+%$^&ksdhkjhjksdadklf&(%^(alkha

Sample output

qj

#include <stdio.h>main(){ int i,n; char c[198]; scanf("%d",&n); while(n--) {  int count=0;  scanf("%s",c);  for(i=0;c[i]!='\0';i++)   if (c[i]>96 && c[i]<123) count++;  if (count%26==0) printf("z\n");  else printf("%c\n",count%26+96); }}