PAT Top 1018 Subnumber 题解

最新推荐文章于 2022-03-16 19:02:12 发布

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本文介绍了一种计算给定正整数N的所有连续非零数字子序列之和的方法，并提供了一个C++代码实现。该算法通过预计算10的幂次来减少计算复杂度，确保了对大数的有效处理。

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Given a positive integer N, let us define a subnumber of N as a consecutive number of digits NOT starting with 0. For example if N=1021, it has 7 subnumbers, namely, 1, 10, 102, 1021, 2, 21 and 1 (again). Here is your task: please calculate the sum of all the subnumbers of N. For 1021, the sum is 1+10+102+1021+2+21+1 = 1158. Since the result may be very large, output the answer modulo 1000000007 (109+7) please.

Input Specification:

Each input file contains one test case, which gives the integer N (0<N<10100000) in a line.

Output Specification:

Print in a line the sum of all N's subnumbers (modulo 1000000007).

对于M位的整数 $d_0d_1d_2d_3 \dots d_{M-1}$ ， $d_j$ 左边包括它自己在内有 $g_j$ 位的非零数字， $d_j$ 对最后结果的贡献为

$contribution(d_j)=(g_j \times d_j\times \sum_{i=0}^{M-j-1}10^i)\%(10^9+7)$ .

最后的结果就是

$result=(\sum_{i=0}^{M-1}contribution(d_i)) \% (10^9+7)$ .

提前以递推的形式计算10的若干次方有助于降低计算复杂度。代码如下。

#include <iostream>
#include <vector>
#include <string>

using namespace std;
const long long N = 1e9 + 7;


int main()
{
    string input;
    cin >> input;

    vector<long long> residual(input.size(), 0);
    residual[0] = 1;
    for(size_t i = 1; i < residual.size(); i++){
        residual[i] = (residual[i - 1] * 10 + 1) % N;
    }

    long long result = 0;
    int g = 0;
    for(size_t i = 0; i < input.size(); i++){
        long long n = input[i] - '0';
        if(n) g += 1;
        int k = input.size() - i;
        result += (g * n * residual[k - 1]);
        result %= N;
    }
    cout << result;
    return 0;
}