PAT甲级1001题解
2017-03-18 20:30
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题目要求如下:
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).
Input
Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.
Output
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.
Sample Input
Sample Output
注:这题是一道简单题,但是前几次提交都拿不到所有测试点,容易有所遗漏。比如1+1,输出002,还有-1+1未输出等低级错误。
(talk is cheap,show me the code.coding......)
#include<stdio.h>
#include<stdlib.h>
int number[10];
int main()
{
int a, b;
bool flag;//用于标记是否为负数
int sum;
while (scanf("%d%d", &a, &b) != EOF)
{
flag = 0;
sum = a + b;
if (sum < 0)
flag = 1;
if (flag == 1)
{
sum = -sum;
printf("-");
}
//int count = 0;
int i = 0;
if (sum == 0)
printf("%d", 0);
bool flag2 = 0;//用于标记是否小于1000
if (sum < 1000)
flag2 = 1;
while (sum)
{
number[i++] = sum % 1000;
sum /= 1000;
}
for (int j = i - 1; j >0; j--)
{
if (j != i - 1)
printf("%03d,", number[j]);
else
printf("%d,", number[j]);
}
if (i >= 1)
{
if (flag2 == 0)
printf("%03d", number[0]);
else
printf("%d", number[0]);
}
printf("\n");
}
return 0;
}
1001. A+B Format (20)
时间限制400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).
Input
Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.
Output
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.
Sample Input
-1000000 9
Sample Output
-999,991
注:这题是一道简单题,但是前几次提交都拿不到所有测试点,容易有所遗漏。比如1+1,输出002,还有-1+1未输出等低级错误。
(talk is cheap,show me the code.coding......)
#include<stdio.h>
#include<stdlib.h>
int number[10];
int main()
{
int a, b;
bool flag;//用于标记是否为负数
int sum;
while (scanf("%d%d", &a, &b) != EOF)
{
flag = 0;
sum = a + b;
if (sum < 0)
flag = 1;
if (flag == 1)
{
sum = -sum;
printf("-");
}
//int count = 0;
int i = 0;
if (sum == 0)
printf("%d", 0);
bool flag2 = 0;//用于标记是否小于1000
if (sum < 1000)
flag2 = 1;
while (sum)
{
number[i++] = sum % 1000;
sum /= 1000;
}
for (int j = i - 1; j >0; j--)
{
if (j != i - 1)
printf("%03d,", number[j]);
else
printf("%d,", number[j]);
}
if (i >= 1)
{
if (flag2 == 0)
printf("%03d", number[0]);
else
printf("%d", number[0]);
}
printf("\n");
}
return 0;
}
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