1073 : Numbers

时间限制Time Limit: 5 秒Sec
内存限制Memory Limit: 512 兆MB
提交次数Submitted: 194 次Times
通过次数Solved: 111 次Times

标准评测 Standard

从标准输入读入，结果写到标准输出。评测将输出拆成 token，与标准答案逐项比较，不按整段逐字节比对。 Read from standard input and write to standard output. The judge splits the output into tokens and compares them with the official answer; it does not compare raw bytes.

比对Compare	token 相同即通过。中间的空格、制表符、换行可多可少。Matching tokens pass. Extra spaces, tabs, or newlines between them are ignored.
不同则错Differs	token 个数或内容不同即错误。23 与 2 3、02 与 2、2.0 与 2 均视为不同。A different token count or value is wrong. 23 vs 2 3, 02 vs 2, and 2.0 vs 2 all differ.
输入Input	标准输入，格式见题面。Standard input; format as in the statement.
输出Output	标准输出。Standard output.
时间与内存Time and memory	记这一份程序的 CPU 时间和内存峰值。题目上的时间限制、内存限制就是这一份的上限。The record is this program’s CPU time and peak memory. The limits on the problem apply to this one run.

模板Template

C++ C Java Python

#include <iostream>
using namespace std;
int main() {
    int a, b;
    cin >> a >> b;
    cout << (a + b) << endl;
    return 0;
}

#include <stdio.h>
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int b = in.nextInt();
        System.out.println(a + b);
    }
}

a, b = map(int, input().split())
print(a + b)

题目描述Description

Bobo has n distinct integers a₁, a₂, …, a_n in [0, 99]. He writes them in decimal notation without leading zeros in a row, obtaning a string s.

Given the string s, find the number of possible array of integers a₁, a₂, …, a_n.

输入格式Input

The input consists of several test cases and is terminated by end-of-file.

Each test case contains a string s.

1 ≤ |s| ≤ 50
There are at most 100 test cases.

输出格式Output

For each test case, print an integer which denotes the result.

样例Sample

出题Author

ftiasch

来源Source

湖南省第十五届大学生计算机程序设计竞赛（HNCPC2019）

题目到此结束End of problem