[C][瘋狂程設][CPE考題20150324] 15A2.UVA10783：Odd Sum

Odd Sum

Given a range [a,b], you are to find the summation of all the odd integers in this range. For example, the summation of all the odd integers in the range [3,9] is 3 + 5 + 7 + 9 = 24.

Input

There can be at multiple test cases. The first line of input gives you the number of test cases,T (1≤ T≤ 100) Then T test cases follow. Each test case consists of 2 integers a and b (0≤ a≤ b≤ 100 )in two separate lines.

Output

For each test case you are to print one line of output - the serial number of the test case followed by the summation of the odd integers in the range [a, b].

Sample Input

2
1
5
3
5

Sample Output

Case 1: 9
Case 2: 8

Notes:

1.儲存輸入的資料

total:表示資料有幾筆，也就是有多少個test case pairs陣列:存放每一筆輸入的起點a與終點b

 int total;
 scanf("%d", &total);

 int input=2*total;
 int pairs[input];
 int i=1,start,end,k,sum;
 
 while(i<=input){
  scanf("%d",&pairs[i]);
  i++;
 }

Code:

#include <stdio.h>

int main(void){

 int total;
 scanf("%d", &total);

 int input=2*total;
 int pairs[input];
 int i=1,start,end,k,sum;
 
 while(i<=input){
  scanf("%d",&pairs[i]);
  i++;
 }
 
 for(i=1;i<=input;i+=2){
  sum=0;
  (pairs[i]%2)?start=pairs[i]:start=pairs[i]+1;
  (pairs[i+1]%2)?end=pairs[i+1]:end=pairs[i+1]-1;
  for(k=start;k<=end;k+=2)
   sum+=k;
  printf("Case %d: %d\n",(i/2+1),sum);
 }
 
 return 0;
}