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