cf468C Hack it!
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
This code will fail only on the test with
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
The first line contains a single integer a (1 ≤ a ≤ 1018).
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
46
1 10
126444381000032
2333333 2333333333333
这题贼鸡儿恶心
令f(x)表示x各个位上数字之和，给个p，要求构造一组[l,r]，使得{f(l)+f(l+1)+...+f(r)} % p = 0
注意到p<=10^18,当x<=10^18时，又有这样的式子成立：f(10^18+x)=f(x)+1 （这很显然，因为就在最前面加了个1）
所以，如果[l,r]表示{f(l)+f(l+1)+...+f(r)} % p
假设[0,10^18] -> a
[1,10^18+1]->a+1
...
[p-a,10^18+p-a]->p (其实这里模p应该是0了)
这个区间就可以了
然后a不会算……就算了个[0,10^18-1]=8.1*10^19，所以右区间还要减一
这个数字好像还是爆long long的，需要81拆成9*9来搞
1 #include<cstdio> 2 int main() 3 { 4 long long p,t,e=1e18;scanf("%lld",&p); 5 t=9*e%p*9%p; 6 printf("%lld %lld ",p-t,e-1+p-t); 7 }