NYOJ 225题 小明求素数积

01.

02.

#include<stdio.h>

03.

#include<string.h>

04.

int

main()

05.

{

06.

bool

n[1000]
= {0};

//标记所有的素数；

07.

int

i, j;

08.

for

(i = 2;
i * i <= 1000; i ++)

09.

{

10.

if

(n[i]
== 0)

11.

for

(j =
i + i; j <= 1000; j += i )

12.

{

13.

n[j] = 1;

14.

}

15.

}

16.

int

T;

17.

scanf

(

"%d"

,
&T);

18.

while

(T --)

19.

{

20.

int

N;

21.

scanf

(

"%d"

,
&N);

22.

int

input[2000]
= {0}, temp[2000] = {0}, output[2000] = {2};

23.

int

num,  
out = 1, k = 0, m = 0;

//in，input中元素个数；out，output中元素的个数；

24.

for

(i =
3; i <= N; i ++)

//模拟手算；

25.

{

26.

int

j
= 0,in = 0;

27.

if

(n[i]
== 0)

//针对所有素数；

28.

{

29.

num = i;

30.

while

(num
!= 0)

31.

{

32.

input[j ++] = num % 10;

33.

num /= 10;

34.

in ++;

35.

}

//将素数一位一位地，倒序地，存到input数组里面；

36.

for

(j
= 0; j < in; j ++)

37.

{

38.

int

l
= j, s[2000] = {0}, c[2000] = {0};

39.

for

(k
= 0; k < out; k ++)

40.

{

41.

s[l] = (c[l] + input[j]
* output[k]) % 10;  

42.

c[l +1] = (c[l] + input[j]
* output[k]) / 10;   

43.

temp[l] += s[l];    

44.

l ++;

45.

}

46.

if

(c[l]
!= 0)

47.

{

48.

temp[l] = c[l];

49.

l = l + 1;

50.

}

51.

m = l;

52.

}

53.

out = m;

54.

int

tempc[2000]
= {0};

55.

for

(m
= 0; m < out; m ++)

56.

{

57.

output[m] = (temp[m] + tempc[m])
% 10;  

58.

tempc[m +1] = (temp[m] + tempc[m])
/ 10;    

59.

}

60.

if

(tempc[m]
!= 0)

61.

{

62.

output[m] = tempc[m];

63.

out = out + 1;

64.

}

65.

memset

(temp,
0,

sizeof

(temp));

//将temp清零；

66.

}

67.

}

68.

if

(out
<= 6)

//总位数<=6, >6, 倒序输出；输出时，竟然还不能以零开头！！！

69.

{

70.

for

(i
= out - 1; i >= 0; i --)

71.

printf

(

"%d"

,
output[i]);

72.

}

73.

else

74.

{

75.

for

(i
= 5; i >= 0; i --)

76.

{

77.

if

(output[i]
!= 0)

78.

break

;

79.

}

80.

for

(j
= i; j >= 0; j --)

81.

printf

(

"%d"

,
output[j]);

82.

}

83.

printf

(

"\n"

);

84.

}

85.

return

0;

86.

}