发表于

给2个有理数，分别计算它们的相加相减相乘相处后的结果，并按要求格式输出。
重点在这个格式化有理数的函数f。

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#include <bits/stdc++.h>
using namespace std; 

long long a, b, c, d;

long long GCD(long long a, long long b) {
    return b == 0 ? abs(a) : GCD(b, a % b);
}

void f(long long a, long long b) {
    if (a == 0) {
        printf("0");
        return;
    }
    if (b == 0) {
        printf("Inf");
        return;
    }
    int flag = (a < 0 && b > 0) || (a > 0 && b < 0);
    a = abs(a), b = abs(b);
    long long integer = a / b;
    if (flag) {
        printf("(-");
    }
    if (integer) {
        printf("%lld", integer);
    }
    if (a % b == 0) {
        if (flag) {
            printf(")");
        }
        return; 
    }
    if (integer != 0) {
        printf(" ");
    }
    long long gcdval = GCD(a, b);
    a = a - integer * b;
    a = a / gcdval, b = b / gcdval;
    printf("%lld/%lld", a, b);
    if (flag) {
        printf(")");
    }
    // printf("%lld/%lld%s", m, n, flag ?  ")" : ""); 也很不错
}

int main() {
    scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
    f(a, b);
    printf(" + ");
    f(c, d);
    printf(" = ");
    f(a * d + b * c, b * d);
    printf("\n");

    f(a, b);
    printf(" - ");
    f(c, d);
    printf(" = ");
    f(a * d - b * c, b * d);
    printf("\n");

    f(a, b);
    printf(" * ");
    f(c, d);
    printf(" = ");
    f(a * c, b * d);
    printf("\n");

    f(a, b);
    printf(" / ");
    f(c, d);
    printf(" = ");
    f(a * d, b * c);
    printf("\n");

    return 0;
}

2/3 -4/2

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

发表于

s2没有出现s1中的字符，并且转换大写后没有出现在答案中（答案不用重复出现相同），把这个字符加入答案。

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#include <bits/stdc++.h>
using namespace std; 

int main() {
    string s1, s2, res;
    cin >> s1 >> s2;
    for (int i = 0; i < s1.length(); i++) {
        if (s2.find(s1[i]) == string::npos && res.find(toupper(s1[i])) == string::npos) {
            res += toupper(s1[i]);
        }
    }
    cout << res << endl;

    return 0;
}

7_This_is_a_test
_hs_s_a_es

7TI

发表于

因为保证每个人的grade不同，所以从高到低排序就可以，不用考虑相同的情况。
不在区间内的，grade赋值-1，往后放，else满足条件的计数cnt++。

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#include <bits/stdc++.h>
using namespace std; 

struct student {
    char name[11];
    char id[11];
    int grade;
};

bool cmp1(student a, student b) {
    return a.grade > b.grade;
}

int main() {
    int n, low, high, cnt = 0;
    scanf("%d", &n);
    vector<student> v(n);
    for (int i = 0; i < n; i++) {
         scanf("%s %s %d", &v[i].name, &v[i].id, &v[i].grade);
    }
    scanf("%d %d", &low, &high);
    for (int i = 0; i < n; i++) {
        if (v[i].grade < low || v[i].grade > high) {
            v[i].grade = -1;
        }
        else {
            cnt++;
        }
    }
    sort(v.begin(), v.end(), cmp1);
    if (cnt == 0) printf("NONE");
    else {
        for (int i = 0; i < cnt; i++) {
            printf("%s %s\n", v[i].name, v[i].id);
        }
    }

    return 0;
}

发表于

每计算一次都计算分子分母的最大公约数，再约分。

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#include <bits/stdc++.h>
using namespace std; 

long long gcd(long long a, long long b) {
    return b == 0 ? abs(a) : gcd(b, a % b);
}

int main() {
    int n;
    long long a, b, suma = 0, sumb = 1, GCD;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%lld/%lld", &a, &b);
        GCD = gcd(a, b);
        a = a / GCD;
        b = b / GCD;
        suma = suma * b + a * sumb;
        sumb = sumb * b;
        GCD = gcd(suma, sumb);
        suma = suma / GCD;
        sumb = sumb / GCD;
    }
    long long integer = suma / sumb;
    suma = suma - integer * sumb;
    if (integer != 0) {
        printf("%lld", integer);
        if (suma != 0) {
            printf(" ");
        }
    }
    if (suma != 0) {
        printf("%lld/%lld", suma, sumb);
    }
    if (integer == 0 && suma == 0) {
        printf("0");
    }

    return 0;
}

发表于

1079 Total Sales of Supply Chain (25分)

pow()函数:求x的y次方的值。
叶子节点为零售商，data保存销售商品的数量，child向量保存子节点。
DFS深度搜索查找叶子节点，累加销售额。
最后再乘以价格p。

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#include <bits/stdc++.h>
using namespace std;

double res = 0.0, p, r;

struct node {
    int data;
    vector<int> child;
};

vector<node> v;

void DFS(int index, int depth) {
    if (v[index].child.size() == 0) {
        res += v[index].data * pow(1 + r, depth);
        return;
    }
    else {
        for (int i = 0; i < v[index].child.size(); i++) {
            DFS(v[index].child[i], depth + 1);
        }
    }
}

int main() {
    int n, k, c;
    scanf("%d %lf %lf", &n, &p, &r);
    r = r / 100;
    v.resize(n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &k);
        if (k == 0) {
            scanf("%d", &v[i].data);
        }
        else {
            for (int j = 0; j < k; j++) {
                scanf("%d", &c);
                v[i].child.push_back(c);
            }
        }
    }
    DFS(0, 0);
    printf("%.1f", res * p);

    return 0;
}