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Re:帮忙看看,C++ AC g++WA 用的STL priorityqueueIn Reply To:Re:帮忙看看,C++ AC g++WA 用的STL priorityqueue Posted by:weakY at 2009-08-23 10:15:23 囧,贴成以前的代码了
现在知道了,g++ 输出double一定要用printf("%f")下面这个应该g++c++都能AC了
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
char s[1000];
int a[300];
struct node
{
int v,w;
} b[10000];
struct poi
{
int p;
};
bool operator<(const poi &A,const poi&B)
{
return b[A.p].v>b[B.p].v;
}
int main()
{
while (scanf("%s",s)==1&&strcmp(s,"END")!=0)
{
int l,i,j;
l=strlen(s);
memset(a,0,sizeof a);
for (i=0;i<l;i++) a[s[i]]++;
priority_queue<poi> Q;
for (j=i=0;i<300;i++)
{
if (a[i]>0)
{
b[j].v=a[i];
b[j].w=0;
poi t;
t.p=j++;
Q.push(t);
}
}
if (j==1)
{
printf("%d %d 8.0\n",b[0].v*8,b[0].v);
continue;
}
while (1)
{
poi t1=Q.top();
Q.pop();
if (Q.empty()) break;
poi t2=Q.top();
Q.pop();
b[j].v=b[t1.p].v+b[t2.p].v;
b[j].w=b[t1.p].w+b[t2.p].w+b[j].v;
poi t;
t.p=j++;
Q.push(t);
}
printf("%d %d %.1f\n",l*8,b[j-1].w,double(l*8/b[j-1].w));
}
}
> printf("%d %d %.1lf\n",l*8,b[j-1].w,(double)(l*8/b[j-1].w));
>
> 你C++能AC?(double)(l*8/b[j-1].w)应该是(double)l*8/b[j-1].w,你的double强制类型转换没有将它们转换成double的除法
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