AtCoder Beginner Contest 430 (A~E)

A:Candy Cookie Law

签到题。

a,b,c,d = map(int,input().split())
if c>=a:
    if d>=b:
        print("No")
    else:
        print("Yes")
else:
    print("No")

B:Count Subgrid

数据范围很小，直接枚举左上角，然后做加法，写起来快。

要想跑得快，甚至可以加上矩阵前缀和，但是没必要。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
#include <unordered_set>
using namespace std;
const int N = 14;
string g[N];
unordered_set<string> S;
int main() {
    int n,m;
    cin>>n>>m;
    int ans=0;
    // cout<<n<<m<<endl;
    for(int i=0;i<n;i++) cin>>g[i];
    for(int i=0;i<n-m+1;i++){
        for(int j=0;j<n-m+1;j++){
            string t="";
            for(int x=i;x<i+m;x++){
                for(int y=j;y<j+m;y++){
                    t+=g[x][y];
                }
            }
            // cout<<t<<endl;
            if(S.count(t)==0){
                S.insert(t);
                ans++;
            }
        }
    }

    cout << ans << "\n";
    return 0;
}

C:Truck Driver

将a和b的位置分组，枚举l，判断r是否合法。

n,a,b = map(int,input().split())
s=input()
posa,posb=[],[]
for i,c in enumerate(s):
    posa.append(i+1) if c=='a' else posb.append(i+1)
ia,ib=0,0
ans=0
for l in range(1,n+1):
    while ia<len(posa) and posa[ia]<l:
        ia+=1
    while ib<len(posb) and posb[ib]<l:
        ib+=1
    rA=posa[ia+a-1] if ia+a-1<len(posa) else n+1
    rB=posb[ib+b-1] if ib+b-1<len(posb) else n+1
    if rA<=n:
        ans+=max(0,rB-rA)
print(ans)

D:Neighbor Distance

每次删除只会影响有限的几个值，考虑清楚，遍历即可。学到了set上使用prev和next函数。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int main() {
    int n;
    cin >> n;
    set<LL> S;
    S.insert(0);
    LL t;
    cin >> t;
    S.insert(t);
    LL ans = 2LL * t;
    cout << ans << endl;
    const LL INF = (1LL<<60);
    for (int i = 0; i < n - 1; ++i){
        cin >> t;
        auto it = S.lower_bound(t);

        if (it == S.begin()){
            LL R = *it;
            LL nextR = (next(it) == S.end() ? INF : *next(it));
            LL oldR = (nextR == INF ? INF : nextR - R);
            LL newR = min((nextR == INF ? INF : nextR - R), R - t);
            LL dt  = R - t;
            ans += (newR + dt) - oldR;
        }else if (it == S.end()){ 
            auto itL = prev(it); 
            LL L = *itL;
            LL prevL = *prev(itL);
            LL oldL = L - prevL;
            LL newL = min(L - prevL, t - L);
            LL dt  = t - L;

            ans += (newL + dt) - oldL;
        }else { 
            LL R = *it;
            auto itL = prev(it);
            LL L = *itL;
            LL prevL = (itL == S.begin() ? INF : *prev(itL));
            LL nextR = (next(it) == S.end() ? INF : *next(it));
            LL oldL = min((prevL == INF ? INF : L - prevL), R - L);
            LL oldR = min((nextR == INF ? INF : nextR - R), R - L);
            LL newL = min((prevL == INF ? INF : L - prevL), t - L);
            LL newR = min((nextR == INF ? INF : nextR - R), R - t);
            LL dt  = min(t - L, R - t);
            ans += (newL + newR + dt) - (oldL + oldR);
        }
        S.insert(t);
        cout << ans << endl;
    }
    return 0;
}

E:Shift String

KMP板子题。

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2e6+10;
char p[N],s[N];
int ne[N];
int main()
{
    int t;
    cin>>t;
    while(t--){
        // memset(ne,0,sizeof ne);
        scanf("%s %s",s+1,p+1);
        int n=strlen(s+1);
        for(int i=2,j=0;i<=n;i++){
            while(j&&p[i]!=p[j+1]) j=ne[j];
            if(p[i]==p[j+1]) j++;
            ne[i]=j;
        }
        for(int i=n+1;i<=2*n;i++){
            s[i]=s[i-n];
        }
        s[2*n+1]='\0';
        // printf("%s \n%s\n",p+1,s+1);
        int flag=false;
        for(int i=1,j=0;i<=2*n;i++){
            while(j&&s[i]!=p[j+1]) j=ne[j];
            if(s[i]==p[j+1]) j++;
            if(j==n){
                printf("%d\n",i-j);
                flag=true;
                break;
            }
        }
        if(!flag)
            printf("-1\n");
    }
    return 0;
}