我正在尝试解决SPOJ问题“Cricket Tournament”。我在python和c中编写了代码。在python中,输入0.0 0/0 300需要大约2秒。但是在C中它会永远运行。 C中的代码正在运行一些较小的测试用例,如19.5 0/0 1
#include<stdio.h>
float ans[10][120][300]={0};
float recursion(int balls, int reqRuns, int wickets);
int readScore(void);
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(ans,0,sizeof(ans));
float overs;
int myruns,wickets,target;
scanf("%f",&overs);
myruns=readScore();
scanf("%d",&wickets);
//printf("%d %d\n",myruns,wickets );
scanf("%d",&target);
//printf("%d %d %d\n",myruns,wickets,target);
if(myruns>=target)
{
printf("%s\n","100.00");
continue;
}
else if(wickets>=10)
{
printf("%s\n", "0.00");
continue;
}
printf("overs = %f\n", overs);
int ov = (int) overs;
int ball = (int)(overs*10)%10;
int totballs = 6*ov+ball;
//printf("%d %d\n",ov,ball );
//printf("%d %d %d\n",totballs, target- myruns,wickets );
float finalAns = recursion(totballs,target-myruns, wickets)*100;
printf("%.2f\n",finalAns);
}
return 0;
}
int readScore()
{
char ch;
int ans2=0;
ch = getchar();
//ch = getchar();
//ans = ans*10 + ch-'0';
//printf("sadasdas %d\n",ch );
while(ch!='/')
{
ch=getchar();
//printf(" ch = %d\n", ch-'0');
if(ch!='/')
ans2 = ans2*10 + ch-'0';
}
//printf("%d\n",ans );
return ans2;
}
float recursion(int balls, int reqRuns, int wickets)
{
if (reqRuns<=0)
return 1;
if (balls==120||wickets==10)
return 0;
if(ans[wickets][balls][reqRuns]!=0)
return ans[wickets][balls][reqRuns];
ans[wickets][balls][reqRuns] = (recursion(balls+1, reqRuns,wickets)+recursion(balls+1, reqRuns-1,wickets)+
recursion(balls+1, reqRuns-2,wickets)+recursion(balls+1, reqRuns-3,wickets)+
recursion(balls+1, reqRuns-4,wickets)+recursion(balls+1, reqRuns-5,wickets)+
recursion(balls+1, reqRuns-6,wickets)+recursion(balls+1, reqRuns,wickets+1)+
2*recursion(balls, reqRuns-1,wickets))/10;
return ans[wickets][balls][reqRuns];
}
from __future__ import division
saved = {}
t = input()
def func(f):
if f in saved: return saved[f]
x,y,z,n = f
if z >= n: return 1
if x == 120: return 0
if y == 10: return 0
saved[f] = (func((x+1,y+1,z,n)) + func((x+1, y,z,n)) + func((x+1,y,z+1,n)) + func((x+1, y, z+2,n)) + func((x+1, y, z+3,n)) + func((x+1, y, z+4,n)) + func((x+1, y, z+5,n))+ func((x+1, y, z+6,n))+ func((x,y,z+1,n)) + func((x,y,z+1,n))) / 10
return saved[f]
def converter(f):
v = f.index('.')
x,y = int(f[:v]), int(f[-1])
return x*6+(y)
for i in range(t):
x,y,z = raw_input().split()
v = y.index('/')
q = int(y[:v])
x,y,z = converter(x), int(y[(v+1):]), int(z)
print '%.2f' % (100 * func((x,y,q,z)))
答案 0 :(得分:3)
你的问题是很多递归的结果是0,所以
if(ans[wickets][balls][reqRuns]!=0)
return ans[wickets][balls][reqRuns];
在许多情况下,无法返回缓存的结果,因此您重新计算许多结果,而Python中的检查f in saved
会阻止重新计算相同的值。
我更改了您的C代码,将ans
的初始条目设置为包含负数(如果您知道您的平台的浮点表示为IEEE754,只需更改为memset(ans, 0x80, sizeof ans);
即可),并用
if (ans[wickets][balls][reqRuns] >= 0)
立即得到结果:
$ time ./a.out < spoj_inp.txt
overs = 0.000000
18.03
real 0m0.023s
user 0m0.020s
sys 0m0.002s
答案 1 :(得分:0)
问题在于您使用scanf。它将空格或换行视为输入的终止符。您很可能在每次输入后输入enter。但问题是它将\ n留在缓冲区中并传递给下一个输入。
如果你没有使用严格的c,你可以打电话
cin.ignore()
。我在你的代码上尝试了它并且能够获得成功的输出。
或者,您可以致电
fflush(stdin);
这也可能有用
答案 2 :(得分:0)
我猜这次递归是值得责备的。代码适用于较小的目标。如果可能的话,摆脱递归。
目标较小:
输入
2
0.0 0/1 10
0.0 2/2 20
输出
100.00
99.99