#include<iostream>
#include<cstdio>
#include<vector>
#include<map>
#define FOR(i,x,y) for(i=x;i<=y;++i)
#define maxa 1000+10
using namespace std;
int a[10];
int main()
{
    string s;
    cin>>s;
    int len = s.length();
    if((s[s.length()-1]-'0')%2==0)
        a[1] = 1;
    else
      a[1] = 0;
      int sum =0,i;
    for(i=0;i<len;++i)
        sum+=s[i]-'0';
    if(sum%3==0)
        a[2] = 1;
    else
        a[2] = 0;
   if(((s[len-2]-'0')*10+s[len-1]-'0')%4==0)
        a[3] = 1;
    else
      a[3] = 0;
   if(((s[len-3]-'0')*100+(s[len-2]-'0')*10+s[len-1]-'0')%8==0)
        a[4] = 1;
    else
      a[4] = 0;

    if(s[len-1]=='1'||s[len-1]=='4'||s[len-1]=='6'||s[len-1]=='9'||s[len-1]=='0')
        a[5] = 1;
    else
          a[5] = 0;
      for(i=1;i<=5;++i)
        cout<<a[i]<<endl;
    return 0;
}

s = input()

print(1 - int(s[-1]) % 2)
a3 = 0
for d in s:
    a3 += int(d)
print(1 - (a3 % 3 != 0))
print(1 - (int(s[-2:]) % 4 != 0))
print(1 - (int(s[-3:]) % 8 != 0))
print(1 - (s[-1] == '2' or s[-1] == '3' or s[-1] == '7' or s[-1] == '8'))

#include <stdio.h>
#include <math.h>
#include <string.h>
int main ()
{
char c[100000];
scanf("%s",c);
int len=strlen(c);
int a=c[len-1]-'0';
int b=(c[len-2]-'0')*10+a;
int d=(c[len-3]-'0')*10+b;
int e=0,i;
for(i=0;i<len;i++)
e+=c[i];
if (a%2==0)
printf("1\n");
else
printf("0\n");
if (e%3==0)
printf("1\n");
else
printf("0\n");
if (b%4==0)
printf("1\n");
else
printf("0\n");
if (d%8==0)
printf("1\n");
else
printf("0\n");
if (a!=2&&a!=3&&a!=7&&a!=8)
printf("1\n");
else
printf("0\n");
}

**
末位判断2的倍数
各个位之和判断3的倍数
末两位判断4的倍数
末三位判断5的倍数
末位014569判断完全平方数
**

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
string s;
int x;
int main(){
    getline(cin,s);
    for(unsigned i = 0;i < s.length();i++) x += s[i] - '0';
    if(s[s.length() - 1] & 1) printf("0\n");
    else printf("1\n");
    if(x % 3 == 0)printf("1\n");
    else printf("0\n");
    if((s[s.length() - 1] - '0' + (s[s.length() - 2] - '0') * 10) % 4 == 0) printf("1\n");
    else printf("0\n");
    if((s[s.length() - 1] - '0' + (s[s.length() - 2] - '0') * 10 + (s[s.length() - 3] - '0') * 100) % 8 == 0) printf("1\n");
    else printf("0\n");
    if(s[s.length() - 1] == '0' || s[s.length() - 1] == '1' || s[s.length() - 1] == '4' || s[s.length() - 1] == '5' || s[s.length() - 1] == '6' || s[s.length() - 1] == '9') printf("1\n");
    else printf("0\n");
    return 0;
}

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我操。。。。。数组一定要开大。。。rte 两次 百度发现开大数组 然后ac

#include<iostream>
#include<cstring>
using namespace std;
char a[1000];
int ein=1,zwei=0;
long int gw=0;
int main()
{
cin>>a;
int len=strlen(a);
if((a[len-1]-'0')%2==0)
cout<<ein;
else
cout<<zwei;
cout<<endl;
for(int i=0;i<len;i++)
gw+=a[i]-'0';
if(gw%3==0)
cout<<ein;
else
cout<<zwei;
cout<<endl;
if(((a[len-2]-'0')*10+a[len-1]-'0')%4==0)
cout<<ein;
else
cout<<zwei;
cout<<endl;
if(((a[len-3]-'0')*100+(a[len-2]-'0')*10+a[len-1]-'0')%8==0)
cout<<ein;
else
cout<<zwei;
cout<<endl;
if(a[len-1]=='1'||a[len-1]=='4'||a[len-1]=='6'||a[len-1]=='9'||a[len-1]=='0')
cout<<ein;
else
cout<<zwei;
return 0;
}

#include<iostream>
#include<string>
#include<set>
using namespace std;

int main() {
string s; cin>>s;
cout<<!(int(s[s.size()-1]-'0')%2)<<endl;
int sum=0;
for (int i=0;i<s.size();++i) sum+=s[i]-'0';
cout<<!(sum%3)<<endl;
sum=0;
for (int i=(s.size()-2<0?0:s.size()-2);i<s.size();++i) sum=sum*10+(s[i]-'0');
cout<<(!(sum%4))<<endl;
for (int i=(s.size()-3<0?0:s.size()-3);i<s.size();++i) sum=sum*10+(s[i]-'0');
cout<<(!(sum%8))<<endl;
set<char> p={'2','3','7','8'};
cout<<(!(p.count(s[s.size()-1])));
return 0;
}

#include<cstdio>
#include<cstring>
int main()
{
char a[1000];
int i,len,temp=0;
gets(a);
len=strlen(a);

if( (a[len-1]-'0') %2==0 )
printf("1\n");
else printf("0\n");

for(i=0;i<len;i++)
{
temp=temp+a[i]-'0';
}
if(temp%3==0)

printf("1\n");
else printf("0\n");

if( ( (a[len-2]-'0') *10 + a[len-1]-'0')%4==0 )
printf("1\n");
else printf("0\n");

if( ( (a[len-3]-'0') *100 + (a[len-2]-'0') *10 + a[len-1]-'0')%8==0 )
printf("1\n");
else printf("0\n");

if( a[len-1]=='1' || a[len-1]=='4' || a[len-1]=='6' || a[len-1]=='9' || a[len-1]=='0' )
printf("1");
else printf("0");

return 0;
}

为什么开35不行 破题 应该行啊

#include<stdio.h>
#include<string.h>
int main()
{
char s[100003];
int l,i,n[100030],f,sum,a4,a8;
while(~scanf("%s",s))
{
f=0,sum=0,a4=a8=0;
l=strlen(s);
for(i=0;i<l;i++)
{
n[i]=s[i]-'0';
a4=a4*10+n[i];
a4%=4;
a8=a8*10+n[i];
a8%=8;
sum+=n[i];
}
if(n[l-1]%2==0)
{
printf("1\n");
f++;
}
else
printf("0\n");
if(sum%3==0)
{
printf("1\n");
f++;
}
else
printf("0\n");
if(a4==0)
{
printf("1\n");
f++;
}
else
printf("0\n");
if(a8==0)
{
printf("1\n");
f++;
}
else
printf("0\n");
if(n[l-1]==0||n[l-1]==1||n[l-1]==4||n[l-1]==5||n[l-1]==6||n[l-1]==9)
printf("1\n");
else
printf("0\n");
}
return 0;
}

首先，看了别人的题解，我知道了被4整除只需要看后2位，因为100肯定被4整除；
被8整除只需要看后3位，因为1000肯定被8整除。
而我做时却是用除法求余数。
其次，完全平方数个位必然是0,1,4,5,6,9.而这道题不需要考虑这个，题的难度对于不知道此规律的人就减小了。
而对于知道此规律的人难度反而加大了。
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;

int a[10003];
int size;
int div(int k){
int i;
int d = 0;
for (i = size; i >= 0; i--){
d *= 10;
d += a[i];
d %= k;
}
return d;
}
int main(){
freopen("in.txt", "r", stdin);

char s[10003];
cin >> s;
int i;
for (i = 0; s[i]; i++);
size = i;
for (i--; i >= 0; i--)a[size - i - 1] = s[i] - '0';
if (a[0] % 2==0)cout << 1 << endl;
else cout << 0 << endl;
int sum = 0;
for (i = 0; i < size; i++)sum += a[i];
if (sum % 3 == 0)cout << 1 << endl;
else cout << 0 << endl;
if (div(4) != 0)cout << 0 << endl;
else cout << 1 << endl;
if (div(8) != 0)cout << 0 << endl;
else cout << 1 << endl;
cout << 1 << endl;
return 0;
}

这题简直逗，f[5]恒为1……
var a:array[1..10000] of longint;
f:array[1..5] of longint;
n,k,i,j,p,q:longint;
s:ansistring;
begin
read(s);
n:=length(s);
for i:=1 to n do
a[i]:=ord(s[n-i+1])-ord('0');
if a[1] mod 2=0 then f[1]:=1 else f[1]:=0;
k:=0;
for i:=1 to n do
begin
k:=k+a[i];
end;
if k mod 3=0 then f[2]:=1 else f[2]:=0;
p:=a[1]+10*a[2];
if p mod 4=0 then f[3]:=1 else f[3]:=0;
q:=a[1]+10*a[2]+100*a[3];
if q mod 8 =0 then f[4]:=1 else f[4]:=0;

f[5]:=1;
for i:=1 to 5 do
writeln(f[i]);
end.

坑题
s[1] in [0,1,4,5,6,9]来判断完全平方
program p1490;
var a,b:array[0..31000] of longint;
n,i,k,t,t1,t2:longint;
s:ansistring;
//
function make(p:longint):longint;
var i:longint;
begin
a:=b;
for i:=a[0] downto 2 do
begin
a[i-1]:=a[i-1]+10*(a[i] mod p);
a[i]:=a[i] div p;
end;
if a[1] mod p=0 then exit(1)
else exit(0);
end;
//
begin
readln(s);
a[0]:=length(s);
for i:=1 to length(s) do a[i]:=ord(s[i])-ord('0');
for i:=1 to a[0] div 2 do begin k:=a[i];a[i]:=a[a[0]+1-i];a[a[0]+1-i]:=k;end;
b:=a;
t:=make(2);t1:=t;writeln(t);
t:=make(3);writeln(t);
t:=make(4);t2:=t;writeln(t);
t:=make(8);writeln(t);
if b[1] in [0,1,4,5,6,9] then write(1)
else write(0);
end.

#include <stdio.h>
#include <string.h>

char s[33];

void dv(int x)
{
printf("%d\n" , x);
}

int main()
{
scanf("%s" , s);

int ilen = strlen(s) , i , sum = 0;

if(s[ilen - 1] % 2 == 0) dv(1);
else dv(0);
for(i = 0 ; i < ilen ; i ++ )
{
sum += s[i] - '0';
}
if(sum % 3 == 0) dv(1);
else dv(0);
if((s[ilen - 1] - '0' + (s[ilen - 2] - '0') * 10) % 4 == 0) dv(1);
else dv(0);
if((s[ilen - 1] - '0' + (s[ilen - 2] - '0') * 10 + (s[ilen - 3] - '0') * 100) % 8 == 0) dv(1);
else dv(0);
if(s[ilen - 1] == '0' || s[ilen - 1] == '1' || s[ilen - 1] == '4' || s[ilen - 1] == '9' || s[ilen - 1] == '6' || s[ilen - 1] == '5') dv(1);
else dv(0);
return 0;
}
如何秒杀？

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 436 KiB, score = 20
测试数据 #1: Accepted, time = 0 ms, mem = 436 KiB, score = 20
测试数据 #2: Accepted, time = 15 ms, mem = 436 KiB, score = 20
测试数据 #3: Accepted, time = 0 ms, mem = 440 KiB, score = 20
测试数据 #4: Accepted, time = 0 ms, mem = 440 KiB, score = 20
Accepted, time = 15 ms, mem = 440 KiB, score = 100

AC的第90题

BS出题人

我开1..32的数组和string怎么改都过不去，最后一怒之下开了个1..10000的数组和ansistring结果AC了

再写题的小心了！

{

ID:darkgod-z

PROG:vijos P1490

HANG:Pascal

}

var

a:array [1..10000] of integer;

i,sum,n:integer;

s:ansistring;

begin

readln(s);

n:=length(s);

for i:=1 to n do

a[i]:=ord(s[n-i+1])-ord('0');

if a[1] mod 2=0 then writeln('1') else writeln('0');

sum:=0;

for i:=1 to n do inc(sum,a[i]);

if sum mod 3=0 then writeln('1') else writeln('0');

if (a[2]*10+a[1]) mod 4=0 then writeln('1') else writeln('0');

if (a[3]*100+a[2]*10+a[1]) mod 8=0 then writeln('1') else writeln('0');

writeln('1');

end.

强烈质疑这道题的正确性……

纪念我的AC100题...

那个可能的确Nice...

可能，无限可能...

晒晒~~

var i,sum:longint;

s:string;

a:array[1..10000] of longint;

procedure reed;

var i,l:longint;

begin

readln(s);

l:=length(s);

sum:=0;

for i:=1 to l do

begin

a[i]:=ord(s[i])-ord('0');

sum:=sum+a[i];

end;

if a[l] mod 2 =0 then writeln(1) else writeln(0);

if sum mod 3 =0 then writeln(1) else writeln(0);

if (a[l-1]*10+a[l]) mod 4 =0 then writeln(1) else writeln(0);

if (a[l-2]*100+a[l-1]*10+a[l]) mod 8=0 then writeln(1) else writeln(0);

writeln(1);

end;

begin

reed;

end.