google在线测试练习题1
2014-10-12 11:17
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Problem
You receive a credit
all available items. From this list you would like to buy two items that add up to the entire value of the credit. The solution you provide will consist of the two integers indicating the positions of the items in your list (smaller number first).
Input
The first line of input gives the number of cases, N. N test cases follow. For each test case there will be:
One line containing the value C, the amount of credit you have at the store.
One line containing the value I, the number of items in the store.
One line containing a space separated list of I integers. Each integer P indicates the price of an item in the store.
Each test case will have exactly one solution.
Output
For each test case, output one line containing "Case #x: " followed by the indices of the two items whose price adds up to the store credit. The lower index should be output first.
Limits
5 ≤ C ≤ 1000
1 ≤ P ≤ 1000
Small dataset
N = 10
3 ≤ I ≤ 100
Large dataset
N = 50
3 ≤ I ≤ 2000
从文件输入,输出到文件中。代码:
You receive a credit
Cat a local store and would like to buy two items. You first walk through the store and create a list
Lof
all available items. From this list you would like to buy two items that add up to the entire value of the credit. The solution you provide will consist of the two integers indicating the positions of the items in your list (smaller number first).
Input
The first line of input gives the number of cases, N. N test cases follow. For each test case there will be:
One line containing the value C, the amount of credit you have at the store.
One line containing the value I, the number of items in the store.
One line containing a space separated list of I integers. Each integer P indicates the price of an item in the store.
Each test case will have exactly one solution.
Output
For each test case, output one line containing "Case #x: " followed by the indices of the two items whose price adds up to the store credit. The lower index should be output first.
Limits
5 ≤ C ≤ 1000
1 ≤ P ≤ 1000
Small dataset
N = 10
3 ≤ I ≤ 100
Large dataset
N = 50
3 ≤ I ≤ 2000
从文件输入,输出到文件中。代码:
#include<iostream> #include<fstream> using namespace std; void quick_sort(int array[],int indexs[], int begin, int end) { if(end > begin) { int pivot = begin; int last_small = begin; int i = end; while(last_small != i) { if(array[i] <= array[pivot]) { int temp = array[i]; int tmp = indexs[i]; array[i] = array[++last_small]; array[last_small] = temp; indexs[i] = indexs[last_small]; indexs[last_small] = tmp; } else i--; } int tmp = array[pivot]; array[pivot] = array[last_small]; array[last_small] = tmp; int temp = indexs[pivot]; indexs[pivot] = indexs[last_small]; indexs[last_small] = temp; quick_sort(array, indexs, begin, last_small - 1); quick_sort(array, indexs, last_small + 1, end); } } int main() { int sum, n; int array[2005]; int n_case; ifstream file2("A-large-practice.in"); ofstream file1("resulta2.txt"); file2 >> n_case; for(int i_case = 1; i_case <= n_case; i_case++) { file2 >> sum; file2 >> n; for(int i = 0; i < n; i++) file2 >> array[i]; int indexs[2005]; for(int i = 0; i < n; i++) indexs[i] = i + 1; quick_sort(array, indexs, 0, n - 1);//sort the list int front = 0; int back = n - 1; while(true) { if(array[front] + array[back] == sum)//found break; else if(array[front] + array[back] > sum) back--; else front++; } if(indexs[front] > indexs[back]) { int tmp = indexs[front]; indexs[front] = indexs[back]; indexs[back] = tmp; } file1 << "Case #" << i_case << ": "; file1 << indexs[front] << ' ' << indexs[back] << endl; } system("pause"); return 0; }
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