NKOJ 1023: A+B+C+D+... 的挑战 - 编程论坛

#2

aipb20072007-07-19 09:38

我的第一直觉就是用string流，getline每一行分别处理。

不知道是否符合你的要求。

#3

leeco2007-07-19 13:49

1023 Accepted GNU C 0.29k 0ms 672KB 2007-07-19 13:45:05

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
    int n,res;
    char buf[10000],*p;
    while(gets(buf)){
        sscanf(strtok(buf,\" \n\"),\"%d\",&res);
        while(p=strtok(NULL,\" \n\")){
            sscanf(p,\"%d\",&n);
            res+=n;
        }
        printf(\"%d\n\",res);
    }
}

#4

HJin2007-07-19 16:18

以下是引用leeco在2007-7-19 13:49:00的发言：
1023 Accepted GNU C 0.29k 0ms 672KB 2007-07-19 13:45:05

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
    int n,res;
    char buf[10000],*p;
    while(gets(buf)){
        sscanf(strtok(buf,\" \n\"),\"%d\",&res);
        while(p=strtok(NULL,\" \n\")){
            sscanf(p,\"%d\",&n);
            res+=n;
        }
        printf(\"%d\n\",res);
    }
}

very smart soln --- thought there is some obvious solution, but it seems it is not that obvious.

Here is my C++ version (I thought those who are interested in this problem already tried it)

#include <iostream>
using namespace std;

int main()
{
int a, sum=0;

    while( cin>> a )
    {
        sum += a;
        while(cin.peek() == ' ')
            getchar();
        if(cin.peek() == '\n')
        {
            cout<<sum<<endl;
            sum = 0;
        }
    }

return 0;
}

#5

mp3aaa2007-07-20 04:38

#include<conio.h>
main()
{
int c,b=0;
char e;
while(1){while(scanf("%d%c",&c,&e)&&e!='\n')b+=c;printf("%d",b+c);}
}
一看就知道是用缓冲区来做不过他出这个题有什么意思啊？

#6

HJin2007-07-20 05:06

mp3aa:

seems to be a good idea.

tried you soln, does not handle the case in which you have spaces before '\n' well, say

4 5 6 '\n'
1 2 3

#7

mp3aaa2007-07-20 13:54

忘了我先做的while(scanf("%d%c",&c,&e)&&e!='\n')b+=c;printf("%d\n",b+c) 后来加的while 忘了初始值了

#include<conio.h>
#include<stdio.h>
main()
{
int c,b=0;
char e;
while(1){
c=0,b=0;
while(scanf("%d%c",&c,&e)&&e!='\n')b+=c;printf("%d\n",b+c);}
}

[此贴子已经被作者于2007-7-20 13:54:40编辑过]

#8

mp3aaa2007-07-20 13:59

楼上是外国人么

#9

Not2007-07-20 14:21

How about this case.
1000000000000000 121212312313 21212121 0
3 2 2 0 4 5

#10

mp3aaa2007-07-20 14:39

以下是引用Not在2007-7-20 14:21:05的发言：
How about this case.
1000000000000000 121212312313 21212121 0
3 2 2 0 4 5

要是大数相加我还不做了来嫌麻烦！

#11

HJin2007-07-20 15:20

you are not asked to deal with overlfow --- instead the main prolbem is to handle the '\n' in the input stream.

#12

mp3aaa2007-07-20 16:59

以下是引用HJin在2007-7-20 15:20:36的发言：
you are not asked to deal with overlfow --- instead the main prolbem is to handle the '\n' in the input stream.

回车没问题啊？我在TC下和 CFREE下都可以
还有你能不能说中文啊
You can speak Chinese？

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[此贴子已经被作者于2007-7-20 17:03:45编辑过]

#13

HJin2007-07-20 20:03

1 2 3
6
4 5
9
9 10
19
11 13 '\n' (note there are 3 spaces before '\n')

Your program for the final test input results in a dead loop.

#14

mp3aaa2007-07-20 23:30

以下是引用HJin在2007-7-20 20:03:49的发言：
1 2 3
6
4 5
9
9 10
19
11 13 '\n' (note there are 3 spaces before '\n')

Your program for the final test input results in a dead loop.

...不是吧题目只是说有空格分开但没有说最后家上空格啊。。。。
按这样挑毛病的话那毛病太多了

#15

leeco2007-07-21 00:50

回复：（mp3aaa）以下是引用HJin在2007-7-20 20:03:49...

以通过ONLINE JUDGE的测试数据为准。你过了就算你对，否则就是错的。

#16

mp3aaa2007-07-21 02:21

恩我试过了如果我不用死循环的话就过了不过我又加了些代码。。。。

#17

leeco2007-07-21 22:39

回复：（mp3aaa）恩我试过了如果我不用死循环的话就...

能否看一下你的方法，发一个过掉的完整代码看看

#18

mp3aaa2007-07-22 16:44

我这里用了两个代码一个是后来做出来的一个是改进的（虽然改进的不在是死循环但是因为在有些编译器里要输入两个ctrl+z才能退出所以通不过）

#include<string.h>
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
char input[100],*p;
int i=0;
while(gets(input)){
i=0;i+=atoi(strtok(input," "));
while(p=strtok(NULL," "))
i+=atoi(p);
printf("%d\n",i);
}
}

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[此贴子已经被作者于2007-7-22 16:47:08编辑过]

#19

mp3aaa2007-07-22 16:45

#include<stdio.h>
int main()
{
int c,b=0;
char e,k='o';
while(e!=EOF){
c=b=0;
while(scanf("%d",&c)&&printf("%d\b"+!(k==(e=getchar())))&&e==' ')b+=c;printf("%d\n",c+b);}
}