138 条题解
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0
panyushan LV 8 @ 2009-09-24 18:01:25
一次就秒杀了，就是爽，！！水题就是水！
#include
using namespace std;
int n,nk[20000];
int mm, result = 0,last = 1;
void mul(int n)
{
int i, j, total = 0, past = 0;
for(i = 1; i 0)
{
nk[++last] = past % 10;
past /= 10;
}
}
bool work(long long a)
{
int i, j;
if(a < 2)
return false;
if(a == 2)
return true;
for(i = 2;i*i < a; i++)
if(a % i == 0)
return false;
return true;
}
int init()
{
scanf("%d",&n);
}
int jc()
{
int i,j;
for(i = n; i >= 2; i --)
mul(i);
for(i = 1; i 0)
{
result += (nk[i] % 10);
nk[i] /= 10;
}
}
}
int outit()
{
if(work(result))
cout -
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var
i,j,k,l,n:longint;
a:array[1..1000,1..2000] of integer;
begin
readln(n);
fillchar(a,sizeof(a),0);
a[1,2000]:=1;
for i:= 2 to n do
begin
k:=0;
for j:= 2000 downto 1 do
begin
k:=k+a*i;
a:=k mod 10;
k:=k div 10;
end;
end;
k:=0;
for i:= 1 to 2000 do
k:=k+a[n,i];
write(k);
for i:= 2 to trunc(sqrt(k)) do
if k mod i=0 then
begin
write('F');
halt;
end;
write('T');
end.1次AC
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0
program zhjh;
var
n,len,x,i,j:longint;
a:array[1..1000000]of longint;
begin
readln(n);
len:=1;
for i:=1 to 10000 do
a[i]:=1;
for i:=1 to n do begin
x:=0;
for j:=1 to len do begin
a[j]:=a[j]*i+x;
x:=a[j] div 10;
a[j]:=a[j] mod 10;
end;
while x>0 do begin
len:=len+1;
a[len]:=x mod 10;
x:=x div 10;
end;
end;
x:=0;
for i:=len downto 1 do
x:=x+a[i];
for i:=2 to trunc(sqrt(x)) do
if x mod i=0 then begin writeln(x,'F'); exit; end;
writeln(x,'T');
end.
end.
会做阶乘,就会做这个 -
0
var a:array[0..5000]of integer;
i,j,n:integer;s:longint;
k:boolean;
begin
read(n);
for i:=0 to 5000 do
a[i]:=0;
a[1]:=1;
for i:=1 to n do
for j:=1 to 5000 do
begin
a[j]:=a[j]*i;
a[j]:=a[j]+(a[j-1] div 10);
a[j-1]:=a[j-1] mod 10;
end;
s:=0;
for i:=1 to 5000 do
s:=s+a[i];
k:=false;
for i:=2 to trunc(sqrt(s)) do
if (s mod i)=0 then k:=true;
if k=true then write(s,'F');
if k=false then write(s,'T');
end. -
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var
a:array[1..10000]of longint;
ans,i,j,k,n,l,len:longint;
ss:string;
function check(v:longint):boolean;
var i:longint;
begin
for i:=2 to trunc(sqrt(v)) do
if v mod i=0 then exit(false);
exit(true);
end;
procedure cheng(v:longint);
var i:longint;
beginfor i:=1 to len do
a[i]:=a[i]*v;
for i:=1 to len do
begin
a:=a+a[i] div 10000;
a[i]:=a[i] mod 10000;
end;
while a[len]0 do inc(len);
end;
begin
readln(n);
a[1]:=1;len:=1;
for i:=2 to n do cheng(i);
ans:=0;
for i:=1 to len do
begin
str(a[i],ss);
for j:=1 to length(ss) do
begin
val(ss[j],k);
inc(ans,k);
end;
end;
write(ans);
if check(ans) then writeln('T')
else writeln('F');
end. -
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szy_1996725 LV 8 @ 2009-08-21 22:01:58
var y,p,k,n,lenc,len1,len2,x,i,j,t:longint;
s:string;
a,b,c:array[0..1000] of integer;
function zs(q:integer):boolean;
var d:integer;
begin
zs:=true;
if q=1 then begin zs:=false; exit; end;
if q=0 then begin zs:=false; exit; end;
for d:=2 to round(sqrt(q)) do
if q mod d=0 then
begin
zs:=false;
exit;
end;
end;
begin
readln(n);
s:='1';
for k:=1 to n do
begin
fillchar(c,sizeof(c),0);
y:=k;
len1:=length(s);
if y=10) and (k=100) and (k1) do dec(lenc);
s:='';
for i:=lenc downto 1 do s:=s+chr(c[i]+48);
end;
for i:=1 to lenc do
t:=t+c[i];
write(t); if zs(t) then writeln('T')
else writeln('F');
end. -
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编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0msvar
n,i,j,k:longint;
a:array[1..1000000] of integer;
begin
fillchar(a,sizeof(a),0);
a[1]:=1;
k:=1;
readln(n);
for i:=2 to n do
begin
a[1]:=a[1]*i;
for j:=2 to k do
begin
a[j]:=a[j]*i+a[j-1] div 10;
a[j-1]:=a[j-1] mod 10;
end;
while a[k]>=10 do
begin
inc(k);
a[k]:=a[k]+a[k-1] div 10;
a[k-1]:=a[k-1] mod 10;
end;
end;
j:=0;
for i:=1 to k do
j:=j+a[i];
write(j);
if j=2 then
begin
writeln('T');
halt;
end;
k:=round(sqrt(j));
for i:=1 to k do
if j mod i=0 then
begin
writeln('F');
halt;
end;
writeln('T');
end.Flag Accepted
题号 P1200
类型(?) 其它
通过 2214人
提交 4513次
通过率 49%
难度 1提交 讨论 题解
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0
program Project1;
var a,b,c,s,d,i:integer;
t:boolean;
begin
readln(a);
s:=1;
t:=true;
for i:=1 to a do
s:=s*i;
begin
问一下，怎么分解各个位并相加？？？？？
谢谢！
d:=b+c;
end;
for i:=1 to trunc(sqrt(d)) do
if d mod i=0 then t:=false;
write(d);
if t then write('T')else write('F');
readln;
readln;
end. -
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var n,i,j,k,l:longint;
s:string;
a:array[1..100] of integer;
begin
readln(n);
k:=1;
for i:=2 to n do k:=k*i;
str(k,s);
l:=length(s);
for i:=1 to l do a[i]:=ord(s[i])-48;
k:=0;
for i:=1 to l do k:=k+a[i];
write(k);
j:=1;
for i:=2 to trunc(sqrt(k)) do
if k mod i=0 then j:=0;
if j=1 then writeln('T')
else writeln('F');
end. -
0
program lx;
var
n,i,c,y :longint;
a,b :array[1..1000]of longint;
f :array[1..1000]of boolean;
j :longint;
procedure add;
var
i :longint;
begin
i:=1;
while f[i]=true do begin
a[i]:=a[i]+b[i];
if a[i]>9999 then begin
a:=a+(a[i] div 10000);
a[i]:=a[i] mod 10000;
end;
i:=i+1;
end;
fillchar(b,sizeof(b),0);
end;
begin
readln(n);
fillchar(a,sizeof(a),0);
fillchar(f,sizeof(f),false);
i:=1;
a[1]:=1;
f[1]:=true;
while i9999 then begin
b[j+1]:=a[j] div 10000;
a[j]:=a[j] mod 10000;
end;
j:=j+1;
end;
add;
i:=i+1;
end;
for i:=1 to 1000 do begin
if f[i]true then begin c:=i-1; break; end;
end;
y:=0;
for i:=1 to c do begin
for j:=1 to 4 do begin
y:=y+(a[i] mod 10);
a[i]:=a[i] div 10;
end;
end;
write(y,' ');
if (n=2) or (n=5) then write('T')
else write('F');
end.
高精度（必须用万进制，否则输出为零）；
另外，只有当n=2或5时为t。 -
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qyjubriskxp LV 10 @ 2009-08-05 00:06:54
爽！一次编译通过，一次测试通过，一次ac
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Dijkstra zkf LV 8 @ 2009-07-30 17:27:27
type arr=array[0..10000]of longint;
var n,i,sum,cheat:longint;
a:arr;
procedure muti(var a:arr; n:longint);
var i,w:longint;
begin
w:=0;
for i:=1 to a[0] do
begin
w:=w+a[i]*n;
a[i]:=w mod 10000;
w:=w div 10000;
end;
while w>0 do
begin
inc(a[0]);
a[a[0]]:=w mod 10000;
w:=w div 100000;
end;
end;
function prime(n:longint):boolean;
var i:longint;
begin
for i:=2 to trunc(sqrt(n)) do
if n mod i=0 then exit(false);
prime:=true;
end;
begin
readln(n);
fillchar(a,sizeof(a),0);
a[0]:=1;
a[1]:=1;
sum:=0;
for i:=2 to n do muti(a,i);
for i:=1 to a[0] do
begin
while a[i]>0 do
begin
inc(sum,a[i] mod 10);
a[i]:=a[i] div 10;
end;
end;
if prime(sum) then write(sum,'T')
else write(sum,'F');
end.代码要写的干净，标准。
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硬做即可
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms#include
#include
using namespace std;
int a[2000]={0};
main ()
{
a[0]=1;
int s,d=0;
cin>>s;
for (int i=s;i>=2 && d==0;i--)
{
if (i
panyushan
LV 8
@ 2009-09-24 18:01:25
