Testing Round #12 597ABC题解

题目链接: 点击打开链接

A: 给出k, a, b. 问你在[a, b]区间内有多少可以整除k的数.

脑洞题目, 计算此区间长度可以存放k的个数即可.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;typedef long long ll;ll a, b, k;int main(int argc, char const *argv[]){    scanf("%lld%lld%lld", &k, &a, &b);    if(b <= 0) {        a = -a;        b = -b;        swap(a, b);    }    a--;    ll ans = b / k - a / k;    if(a < 0 && a % k) ans++;    printf("%lld\n", ans);    return 0;}

B: 给出n个订单的起始时间和结束时间, 问你最多可以接多少个订单, 注意同一时间点不可存在两订单.

贪心题, 按照结束时间排序, 如果开始时间在当前结束时间之后则ans++.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "utility"using namespace std;const int MAXN = 5e5 + 5;int n, x, y, ans;pair<int, int> p[MAXN];int main(int argc, char const *argv[]){    scanf("%d", &n);    for(int i = 0; i < n; ++i) {        scanf("%d%d", &x, &y);        p[i] = make_pair(y, x);    }    sort(p, p + n);    x = p[0].second, y = 0;    for(int i = 0; i < n; ++i)        if(p[i].second > y) {            x = p[i].second;            y = p[i].first;            ans++;        }    printf("%d\n", ans);    return 0;}

C: 给出n个数, 问你上升子序列长度为k + 1的有多少个.

树状数组题目, query(x, y)返回长度为x, 子序列长度为y的上升子序列个数.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int MAXN = 1e5 + 5;typedef long long ll;int n, k;ll a[12][MAXN];void update(int i, int j, ll v){    while(i <= n + 1) {        a[j][i] += v;        i += i & (-i);    }}ll query(int i, int j){    if(i == 0) return 0;    return a[j][i] + query(i - (i & -i), j);}int main(int argc, char const *argv[]){    scanf("%d%d", &n, &k);    update(1, 0, 1);    for(int i = 0; i < n; ++i) {        int x;        scanf("%d", &x);        x++;        for(int j = k + 1; j > 0; --j)            update(x, j, query(x, j - 1));    }    printf("%lld\n", query(n + 1, k + 1));    return 0;}