用java解决问题。 Given a string, return the sum of the digits 0-9 that appear in the string, ignoring all other characters. Return 0 if there are no digits in the string. (Note: Character.isDigit(char) tests if a char is one of the chars '0', '1', .. '9'. Integer.parseInt(string) converts a string to an int.)

可以使用Java中的正则表达式来实现该问题。具体实现步骤如下：

1.使用正则表达式匹配字符串中的所有数字，可以使用`\\d`表示数字。

2.将匹配到的数字字符串转换为整数类型并累加。

3.如果没有匹配到数字，则返回0。

下面是代码实现：

public int sumDigits(String str) {
    int sum = 0;
    Pattern pattern = Pattern.compile("\\d"); // 正则表达式，匹配数字
    Matcher matcher = pattern.matcher(str);
    while (matcher.find()) { // 查找匹配到的数字，并累加
        sum += Integer.parseInt(matcher.group());
    }
    return sum;
}

使用示例：

String str = "abc123def456";
int sum = sumDigits(str);
System.out.println(sum); // 输出 21，即 1+2+3+4+5+6

相关问题

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number. Now for any pair of positive integers N1​ and N2​, your task is to find the radix of one number while that of the other is given. Input Specification: Each input file contains one test case. Each case occupies a line which contains 4 positive integers: N1 N2 tag radix Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2. Output Specification: For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix. 用c++写代码 并且要考虑radix大于36的情况

您想解决一个关于进制转换的问题，其中需要确定一个数字的进制，而另一个数字的进制已知。您希望用C++编写代码，并且要考虑进制大于36的情况。以下是一个可能的实现：

#include <iostream>
#include <string>
#include <cmath>

using namespace std;

int charToInt(char c) {
    if (isdigit(c)) {
        return c - '0';
    } else {
        return c - 'a' + 10;
    }
}

long long convertToDecimal(string num, int radix) {
    long long result = 0;
    int power = 0;
    for (int i = num.length() - 1; i >= 0; i--) {
        result += charToInt(num[i]) * pow(radix, power);
        power++;
    }
    return result;
}

int findRadix(string num, long long target) {
    char maxDigit = *max_element(num.begin(), num.end());
    int minRadix = charToInt(maxDigit) + 1; // Smallest possible radix
    
    // Binary search to find the radix
    int left = minRadix;
    int right = max(36, minRadix); // Consider radix larger than 36
    while (left <= right) {
        int mid = (left + right) / 2;
        long long converted = convertToDecimal(num, mid);
        if (converted == target) {
            return mid;
        } else if (converted < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return -1; // Impossible
}

int main() {
    string N1, N2;
    int tag, radix;
    cin >> N1 >> N2 >> tag >> radix;
    
    if (tag == 1) {
        long long decimalN2 = convertToDecimal(N2, radix);
        int resultRadix = findRadix(N1, decimalN2);
        if (resultRadix == -1) {
            cout << "Impossible" << endl;
        } else {
            cout << resultRadix << endl;
        }
    } else {
        long long decimalN1 = convertToDecimal(N1, radix);
        int resultRadix = findRadix(N2, decimalN1);
        if (resultRadix == -1) {
            cout << "Impossible" << endl;
        } else {
            cout << resultRadix << endl;
        }
    }
    
    return 0;
}

这个程序首先定义了一个`charToInt`函数，用于将字符转换为对应的整数值。然后，`convertToDecimal`函数将给定进制的数字转换为十进制数。最后，`findRadix`函数使用二分查找来确定另一个数字的进制。在主函数中，根据输入的`tag`值选择相应的计算方法，并输出结果。

希望这可以帮助到您！如果您有任何其他问题，请随时问我。

用c++解决You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

以下是使用C++解决该问题的代码示例：

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 110;
const double INF = 1e9;

struct Edge {
    int from, to;
    double rate, commission;
    Edge(int from, int to, double rate, double commission) : from(from), to(to), rate(rate), commission(commission) {}
};

vector<Edge> edges;
double dist[MAXN];

bool bellman_ford(int n, int m, int s, double v) {
    memset(dist, 0, sizeof(dist));
    dist[s] = v;
    for (int i = 0; i < n - 1; i++) {
        bool updated = false;
        for (int j = 0; j < m; j++) {
            Edge e = edges[j];
            if (dist[e.from] > e.commission && dist[e.from] * e.rate - e.commission > dist[e.to]) {
                dist[e.to] = dist[e.from] * e.rate - e.commission;
                updated = true;
            }
        }
        if (!updated) {
            break;
        }
    }
    for (int j = 0; j < m; j++) {
        Edge e = edges[j];
        if (dist[e.from] > e.commission && dist[e.from] * e.rate - e.commission > dist[e.to]) {
            return true;
        }
    }
    return false;
}

int main() {
    int n, m, s;
    double v;
    scanf("%d%d%d%lf", &n, &m, &s, &v);
    for (int i = 0; i < m; i++) {
        int a, b;
        double rab, cab, rba, cba;
        scanf("%d%d%lf%lf%lf%lf", &a, &b, &rab, &cab, &rba, &cba);
        edges.push_back(Edge(a, b, rab, cab));
        edges.push_back(Edge(b, a, rba, cba));
    }
    if (bellman_ford(n, m * 2, s, v)) {
        printf("YES\n");
    } else {
        printf("NO\n");
    }
    return 0;
}

该代码使用了Bellman-Ford算法来判断是否可以增加资产。具体实现方法为：首先将起始点的资产赋值为v，其余点的资产赋值为0；然后进行n-1轮松弛操作，每次遍历所有边，如果满足条件（即当前点的资产大于手续费，且通过该边操作后能够得到更多的资产），则更新该点的资产；最后再进行一轮遍历，如果仍然满足条件，则说明可以增加资产，否则不能。