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Re:一种解法.In Reply To:Re:又一个把自己代码拿出来炫的 Posted by:I08690127 at 2009-07-27 12:24:01 第一种算法有误,但是能过了北大OJ,不信可以试试
可以试试
3
0 0 -1
0 0 1
-1 0 2
输出结果:3
将第一种算法改为如下就对了:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
int c,res=0,k,i,j;
int d[100][100];
int s[101],a[100];
while(scanf("%d",&c)==1)
{
for( i=0;i<c;i++)
{
for( j=0;j<c;j++)
{
scanf("%d",&d[i][j]);
}
}
for( i=0;i<c;i++)
{
for( j=(i+1);j<c;j++)
{
memset(a,0,sizeof(a));
memset(s,0,sizeof(s));
for(k=0;k<c;k++)
{
int m;
for( m=i;m<=j;m++)
a[k]+=d[m][k];
if(s[k]>=0)
s[k+1]=s[k]+a[k];
else
s[k+1]=a[k];
}
int c;
for( c=0;c<k+1;c++)
{
if(res<s[c])
res=s[c];
}
}
}
printf("%d",res);
}
return 0;
}
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