acm的题目

[Tdreamge]

poj 1007
Language:DefaultDNA Sorting
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 61135 Accepted: 24156

Description

One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be---exactly the reverse of sorted).

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length.

Input

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.
Output

Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. Since two strings can be equally sorted, then output them according to the orginal order.
Sample Input

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
Source


题目意思是说有一些字符串，这些字符串的字符不都是按先后顺序排的。这些字符串都对应一个值，这个值是倒序数的总和。例如：AACEDGG只有E和D是反的，所以这个值是1。GCA这个串的值就是3（GA、GC、CA都是反的）。
#include <stdio.h>
#include <string.h>
int Sum(char a[],int n)
{
int k,j,sum=0;
for (j=0;j<n;j++)
{
for (k=j+1;k<n;k++)
{
if (a[j]>a[k]) sum++;
}
}
return sum;
}
int main(int argc, char *argv[])
{
int n,m;
int i,j,t;
char a[50][110],c[110];
int b[50];
scanf("%d%d",&n,&m);
for (i=0;i<m;i++)
{
scanf("%s",a[i]);
b[i]=Sum(a[i],n); //求b[i]的值
}
//冒泡
for (i=0;i<n-1;i++)
for (j=0;j<n-1-i;j++)
{
if(b[j]>b[j+1])
{
t=b[j]; strcpy(c,a[j]);
b[j]=b[j+1]; strcpy(a[j],a[j+1]);
b[j+1]=t; strcpy(a[j+1],c); //若前个b[i]大的话就交换 连同字符窜一起交换
}
}
for (i=0;i<m;i++)
printf("%s\n",a[i]);
return 0;
}


为什么会有一个显示不出来 把数组c的大小改成10就可以了
谢谢大家

[面包大师]

        for (i=0;i<n;i++)
	{
		memset(a[i], 0, 110);//这儿将字符串置空，省的有乱码
 		scanf("%s",a[i]);   
		b[i]=Sum(a[i],m); 
	}

[面包大师]

        for (i=0;i<m-1;i++)//排序的这个地方要用m，不是n
	{
		for (j=0;j<m-1-i;j++)  
		{
			if(b[j]>b[j+1])
			{
				t=b[j]; strcpy(c,a[j]);
				b[j]=b[j+1]; strcpy(a[j],a[j+1]);   
				b[j+1]=t; strcpy(a[j+1],c); 
			}
		}
	}

[gqjjqg]

#include <stdio.h>
int Sum(char a[],int n)
{
	int k,j,sum=0;
	for (j=0;j<n;j++)
	{
		for (k=j+1;k<n;k++)
		{
			if (a[j]>a[k]) sum++;  
		}  
	}
	return sum;
}
int main()
{
	int n,m;
	int i,j,t;
	char a[128][110],c[110]; //m 是 100以内，你开50肯定不够，n 是50内，倒是够了。
	int b[128];
	//交换地址，避免copy内存
	char *pStrOutList[128];
	char *tp;
	scanf("%d %d",&n,&m);
	for (i=0;i<m;i++)
	{
		scanf("%s", a[i]);   
		b[i] = Sum(a[i],n); //求b[i]的值   
		pStrOutList[i] = a[i];
	}
	// 冒泡
	for (i=0;i<m;i++) //n 和 m 别搞混了
		for (j=i + 1;j<m;j++)   
		{
			if(b[i]>b[j])
			{
				t=b[i]; //strcpy(c,a[j]);
				b[i]=b[j]; //strcpy(a[j],a[j+1]);   
				b[j]=t; //strcpy(a[j+1],c); //若前个b[i]大的话就交换 连同字符窜一起交换   
				
				tp = pStrOutList[i];
				pStrOutList[i] = pStrOutList[j];
				pStrOutList[j] = tp;
			}
		}
	for (i=0;i<m;i++)
		printf("%s\n", pStrOutList[i]);
	return 0;
}

拿你的代码A了一把...