[AtCoderContest075F]Mirrored
[AtCoderContest075F]Mirrored
试题描述
For a positive integer \(n\), we denote the integer obtained by reversing the decimal notation of n (without leading zeroes) by \(rev(n)\). For example, \(rev(123)=321\) and \(rev(4000)=4\).
You are given a positive integer \(D\). How many positive integers \(N\) satisfy \(rev(N)=N+D\)?
\(rev(n)\) 表示 \(n\) 在十进制表示下的倒过来写的数,对于给定的 \(D\),求有多少个正整数 \(N\) 满足 \(rev(N) = N + D\)。
输入
Input is given from Standard Input in the following format:
D
输出
Print the number of the positive integers \(N\) such that \(rev(N)=N+D\).
输入示例1
63
输出示例1
2
输入示例2
75
输出示例2
0
输入示例3
864197532
输出示例3
1920
数据规模及约定
\(D\) is an integer.
\(1 \le D < 10^9\)
题解
考虑从两头向中间依次确定每一位,考虑每一位的贡献。
abcdefg
- gfedcba
所以 \(a - g\) 的贡献是 \((a - g) * 999999\),下一位贡献是 \((b - f) * 999900\)……所以 \(D\) 必须是 \(9\) 的倍数。令 \(d = D \div 9\)
那么每一位的贡献依次是:
111111
11110
1100
上面是偶数位的情况,偶数位类似。
注意到上面的形式,我们可以根据 \(d\) 确定出每一位的数位差是多少。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 15
#define LL long long
int d;
LL ans;
int main() {
d = read();
if(d % 9) return puts("0"), 0;
d /= 9;
// printf("d: %d\n", d);
LL ini = 0, ten = 1;
for(int i = 1; i <= 17; i++) {
LL tmp = 1, D = d;
ini = ini * 10 + 1;
LL base = ini, lten = ten, rten = 1;
for(int j = 0; j <= (i >> 1); j++) {
// printf("%lld * %lld [tmp *= %lld] (%lld)\n", base, D % (rten * 10) / rten, 10 - abs(D % (rten * 10) / rten) - (!j ? 1 : 0), D);
tmp *= 10 - abs(D % (rten * 10) / rten) - (!j ? 1 : 0);
D -= base * (D % (rten * 10) / rten);
base -= lten + rten;
lten /= 10; rten *= 10;
}
// printf("%lld * %d (%lld)\n", base, D % (rten * 10) / rten, D);
ten *= 10;
// printf("(%lld %lld)\n", ini, tmp);
if(!D) ans += tmp;
}
printf("%lld\n", ans);
return 0;
}