題目:輸入一個遞增排序的數組和一個數字s,在數組中查找兩個數,使得它們的和正好是s。如果有多對數字的和等於s,輸出任意一對即可。
思路:最初我們找到數組的第一個數字和最後一個數字。首先定義兩個指針,第一個指針指向數組的第一個(也就是最小的)數字,第二個指針指向數組的最後一個(也就是最大的)數字。當兩個數字的和大於輸入的數字時,把較大的數字往前移動;當兩個數字的和小於數字時,把較小的數字往後移動;當相等時,打完收工。這樣掃描的順序是從數組的兩端向數組的中間掃描。
#include "stdafx.h"
#include<iostream>
bool FindNumbersWithSum(int data[], int length, int sum, int* num1, int* num2)
{
bool found = false;
if(length < 1 || num1 == NULL || num2 == NULL)
return found;
int ahead = length -1;
int behind = 0;
while(ahead > behind)
{
long long curSum = data[ahead] + data[behind];
if(curSum == sum)
{
*num1 = data[behind];
*num2 = data[ahead];
found = true;
break;
}
else if(curSum > sum)
ahead --;
else
behind ++;
}
return found;
}
int main()
{
int data[] = {1,2,4,7,11,15};
int length = sizeof(data)/sizeof(int);
int sum = 15;
int num1, num2;
int result = FindNumbersWithSum(data, length, sum, &num1, &num2);
if(result)
{
if(num1 + num2 == sum)
printf("%d %d\n", num1, num2);
else
printf("Failed.\n");
}
else
printf("Failed.\n");
return 0;
}

題目:輸入一個正數S,打印出所有和為S的連續正數序列(至少有兩個數)。例如輸入15,由於1+2+3+4+5=4+5+6=7+8=15,所以結果打印出3個連續序列1~5,4~6和7~8.
思路:
兩個數small和big分別表示序列的最小值和最大值。首先把small初始化為1,big初始化為2.如果從small到big的序列的和大於S,可以從序列中去掉較小的值,也就是增大small的值。如果從small到big的序列的和小於S,可以增大big,讓這個序列包含更多的數字。因為這個序列至少要有兩個數字,我們一直增加small到(1+S)/2為止。
1 #include "stdafx.h"
2
3 void PrintContinuousSequence(int small, int big);
4
5 void FindContinuousSequence(int sum)
6 {
7 if(sum < 3)
8 return;
9
10 int small = 1;
11 int big = 2;
12 int middle = (1 + sum) / 2;
13 int curSum = small + big;
14
15 while(small < middle)
16 {
17 if(curSum == sum)
18 PrintContinuousSequence(small, big);
19
20 while(curSum > sum && small < middle)
21 {
22 curSum -= small;
23 small ++;
24
25 if(curSum == sum)
26 PrintContinuousSequence(small, big);
27 }
28
29 big ++;
30 curSum += big;
31 }
32 }
33
34 void PrintContinuousSequence(int small, int big)
35 {
36 for(int i = small; i <= big ; i ++)
37 printf("%d ", i);
38
39 printf("\n");
40 }
41
42 int main(int argc, char const *argv[])
43 {
44 int sum = 9;
45 printf("test for %d:\n", sum);
46 FindContinuousSequence(sum);
47
48 sum = 15;
49 printf("test for %d:\n", sum);
50 FindContinuousSequence(sum);
51
52 sum = 100;
53 printf("test for %d:\n", sum);
54 FindContinuousSequence(sum);
55
56 return 0;
57 }
