题目链接
Description
For a positive integer (n), let us define (f(n)) as the number of digits in base (10).
You are given an integer (S(1≤S≤10^8)). Count the number of the pairs of positive integers ((l,r)) ((l≤r)) such that (f(l)+f(l+1)+…+f(r)=S), and find the count modulo (10^9+7).
题解
分两种情况讨论。
(1. f(l)leq 7)
则至多有(10^8/8=12500000)个八位数，所以(rleq 22500000).
因此，可以预处理出(f[1..22500000])，然后采用 尺取法 得到 当(f(l)leq 7)时 的((l,r))组数.
(2. f(l)geq 8)
此时，由 (S) 的限制，易知 (f(r)-f(l)leq 1).
令 (t=r-l+1)，枚举 (t)
若 (f(l)==f(r))
则显然要求 (t|S),
记(N)位数的个数右 (D_N) 个，则符合要求的 ((l,r)) 组数为 (D_{frac{S}{t}}-t+1)
若 (f(r)-f(l)==1)
设 (len=f(l)) 且长度为 (len) 的数字有 (x) 个，长度为 (len+1) 的数字有 (y) 个，则有
[egin{cases}
x+y=t\
x*len+y*(len+1)=S
end{cases}
]
其中(len=lfloorfrac{S}{t}
floor).
记(S=tq+r (0leq rlt t))
则方程组可转化为
[egin{cases}
x+y=t\
xq+y(q+1)=tq+r
end{cases}
]
所以有
[egin{cases}
x=t-r=t-S\%t\
y=r=S\%t
end{cases}
]
即对每个(t)此时有且仅有一组解。
统计上述三部分的答案，即为最终答案。
Code
#include <bits/stdc++.h>
#define lim1 10000000
#define lim2 22500000
using namespace std;
typedef long long LL;
int f[lim2+10];
const LL mod = 1e9+7;
void init() {
int l = 1, r = 10;
for (int k = 1; k <= 7; ++k) {
for (int i = l; i < r; ++i) f[i] = k;
l = r; r *= 10;
}
for (int i = lim1; i <= lim2; ++i) f[i] = 8;
}
LL poww(LL a, LL b) {
LL ret = 1;
while (b) {
if (b & 1) (ret *= a) %= mod;
(a *= a) %= mod;
b >>= 1;
}
return ret;
}
LL count1(LL x) {
int l = 1, r = 1, ret = 0; LL sum = 0;
while (true) {
while (r <= lim2 && sum < x) sum += f[r++];
if (sum < x) break;
if (sum == x) (ret += 1) %= mod;
sum -= f[l++];
if (l == lim1) break;
}
return ret;
}
LL count2(LL x) {
LL upp = x/8,
ret = upp;
for (int t = 1; t <= upp; ++t) {
if (x % t == 0) {
int len = x / t;
LL sub = 9 * poww(10, len-1) % mod;
(sub += mod - t) %= mod;
(ret += sub) %= mod;
}
}
return ret;
}
int main() {
init();
LL n;
scanf("%lld", &n);
LL ans1 = count1(n), ans2 = count2(n);
printf("%lld
", (ans1+ans2)%mod);
return 0;
}