栏目分类:
子分类:
返回
名师互学网用户登录
快速导航关闭
当前搜索
当前分类
子分类
实用工具
热门搜索
名师互学网 > IT > 软件开发 > 后端开发 > C/C++/C#

【CF题解】Educational Codeforces Round 118 (Rated for Div. 2)【A-C】

C/C++/C# 更新时间: 发布时间: IT归档 最新发布 模块sitemap 名妆网 法律咨询 聚返吧 英语巴士网 伯小乐 网商动力

【CF题解】Educational Codeforces Round 118 (Rated for Div. 2)【A-C】

A. Long Comparison 思路
  1. 先判断字符长度
  2. 相等的字符长度判断填满0后判断字典序
AC代码
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")

#include
using namespace std;

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2e5+10;

ll a[maxn];
ll b[maxn];

int main(void){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif 
    int t;
    cin >> t;
    while(t--){
        string x1;
        int p1;
        string x2;
        int p2;
        cin >> x1 >> p1;
        cin >> x2 >> p2;
        int len1 = x1.size();
        int len2 = x2.size();
        if(len1+p1 > len2+p2) cout<<">"< x2) cout<<">"<"<"<
using namespace std;

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2e5+10;

ll a[maxn];
ll b[maxn];

int main(void){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif 
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
        for(int i = 1; i <= n; ++i) cin >> a[i];
        sort(a+1,a+1+n);
        for(int i = 1; i <= n/2; ++i){
            cout< 
C. Poisoned Dagger 
思路 

二分K即可。

AC代码
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")

#include
using namespace std;

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2e5+10;

ll a[maxn];
ll b[maxn];
ll n,h;
ll check(ll mid){
    ll sum = 0;
    for(int i = 1; i <= n-1; ++i) sum += min(mid,a[i+1]-a[i]);
    sum += mid;
    return sum;
}
int main(void){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif 
    int t;
    cin >> t;
    while(t--){
        cin >> n >> h;
        for(int i = 1; i <= n; ++i) cin >> a[i];
        ll l = 1;
        ll r = h;
        while(l < r){
            ll mid = (l+r) >> 1;
            if(check(mid) < h) l = mid+1;
            else r = mid;
        }
        cout<
转载请注明:文章转载自 www.mshxw.com
本文地址:https://www.mshxw.com/it/629219.html
我们一直用心在做
关于我们 文章归档 网站地图 联系我们

版权所有 (c)2021-2022 MSHXW.COM

ICP备案号:晋ICP备2021003244-6号