Codeforces Round #516 (Div. 2, by Moscow Team Olympiad)

题目链接

A. Make a triangle!

题意

让某段最少增加多少使得构成三角形

思路

让较小两段往最长段去凑

代码

#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl;
using namespace std;

int a[5];

int main(){
    scanf("%d%d%d",&a[1],&a[2],&a[3]);
    sort(a+1,a+1+3);
    if(a[1]+a[2] > a[3])puts("0");
    else printf("%d\n",a[3]-a[2]-a[1]+1);
    return 0;
}　

B. Equations of Mathematical Magic

题意

存在多少种$x$使得$x$与$a$满足题式

思路

打表发现与$a$的二进制表示中$1$的数量有关

代码

#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl;
using namespace std;
typedef long long LL;

int t,n;
LL pw[35];

int main(){
    scanf("%d",&t);
    pw[0]=1;
    for(int i=1;i<35;i++)pw[i]=2LL*pw[i-1];
    while(t--){
        scanf("%d",&n);
        int cnt=0;
        for(int i=0;i<=31;i++)if((1<<i)&n)cnt++;
        printf("%lld\n",pw[cnt]);
    }
    return 0;
}

C. Oh Those Palindromes

题意

用原串字符构造出子串中回文串最多的串

思路

让相同字符相邻必定为最优情况

代码

#include <bits/stdc++.h>
#define cerr << #x << " = " << x << endl;
const int maxn = 1e5+5;
using namespace std;

int n,cnt[30];
char s[maxn];

int main(){
    scanf("%d%s",&n,s+1);
    for(int i=1;i<=n;i++)cnt[s[i]-'a']++;
    for(int i=0;i<26;i++){
        for(int j=0;j<cnt[i];j++)printf("%c",i+'a');
    }puts("");
}

D. Labyrinth

题意

从某点出发,左右移动次数有限,最多能到达几个合法点

思路

优先进行上下移动,用双端队列模拟bfs

代码

#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl;
const int maxn = 2e3+5;
using namespace std;

int n,m,sr,sc,lm,rm;
int tot,vis[maxn][maxn];
int tr[4]={1,-1,0,0},tc[4]={0,0,1,-1};
char mp[maxn][maxn];

struct node{
    int r,c,t1,t2;
};

void bfs(){
    deque<node>Q;
    vis[sr][sc]=1;
    Q.push_back({sr,sc,lm,rm});
    while(!Q.empty()){
        node tmp=Q.front();
        Q.pop_front();
        for(int i=0;i<4;i++){
            node now=tmp;
            now.r+=tr[i],now.c+=tc[i];
            if(now.r >= 1 && now.r <= n && now.c >= 1 && now.c <= m && !vis[now.r][now.c] && mp[now.r][now.c] == '.'){
                if(i < 2){tot+=vis[now.r][now.c]=1;Q.push_front({now.r,now.c,now.t1,now.t2});}
                else{
                    if(tc[i] == -1 && now.t1){
                        tot+=vis[now.r][now.c]=1;
                        Q.push_back({now.r,now.c,now.t1-1,now.t2});
                    }
                    if(tc[i] ==  1 && now.t2){
                        tot+=vis[now.r][now.c]=1;
                        Q.push_back({now.r,now.c,now.t1,now.t2-1});
                    }
                }
            }
        }
    }
}

int main(){
    scanf("%d%d%d%d%d%d",&n,&m,&sr,&sc,&lm,&rm);
    for(int i=1;i<=n;i++)scanf("%s",mp[i]+1);
    bfs();
    printf("%d\n",tot+1);
    return 0;
}

E. Dwarves, Hats and Extrasensory Abilities

题意

交互题,每次询问一个点的颜色,最后用一条直线将两种颜色的点分开,不存在这种直线也会WA

思路

先询问坐标上的点,二分枚举剩余点,颜色与第一点相同就$l=mid$,否则$r=mid$,等于把与第一点相同色的点分在一边,剩下的分在另一边

代码

#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl;
const int maxn = 35;
using namespace std;

int n;
string str[maxn];

int main(){
    cin >> n;
    cout << "0 2\n";
    fflush(stdout);
    cin >> str[1];
    int l=1,r=(int)1e9,mid;
    for(int i=2;i<=n;i++){
        mid=(l+r)/2;
        cout << mid << ' ' << 2 << endl;
        fflush(stdout);
        cin >> str[i];
        (str[i][0] == str[1][0]) ? l=mid : r=mid;
    }
    cout << l << ' ' << 3 << ' ' << r << ' ' << 1 << endl;
    fflush(stdout);
    return 0;
}

F. Candies for Children

留坑,寒假补