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  • Codeforces Round #568 Div.2

    A. Ropewalkers

    #include <bits/stdc++.h>
    using namespace std;
     
    int num[5];
    int d;
     
    int main() {
        for(int i = 0; i < 3; i ++)
            scanf("%d", &num[i]);
        scanf("%d", &d);
        
        sort(num, num + 3);
        int ans = 0;
        
        if(num[1] - num[0] < d)
            ans += d - num[1] + num[0];
        if(num[2] - num[1] < d)
            ans += d - num[2] + num[1];
        
        printf("%d
    ", ans);
        
        return 0;
    }
    View Code

    B. Email from Polycarp

    #include <bits/stdc++.h>
    using namespace std;
    int T;
    string s, t;
    
    bool isSubsequence(string s, string t) {
        int ls = s.length(), lt = t.length();
        if(!ls) return true;
        if(!lt && ls) return false;
         
        int pos = 0, temp = 0;
        char c = s[pos];
        while(pos <= ls && temp <= lt) {
            if(s[pos] == t[temp]) {
                c = s[pos];
                pos ++;
            } else {
                if(t[temp] != c) return false;
            }
            temp ++;
        }
        return pos == ls + 1;
    }
    
    int main() {
        scanf("%d", &T);
        while(T --) {
            cin >> s >> t;
            if(isSubsequence(s, t)) printf("YES
    ");
            else printf("NO
    ");
        }
        return 0;
    }
    View Code

    C1. Exam in BerSU (easy version)

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 2e5 + 10;
    int N, M;
    int a[maxn];
    long long num[110];
    
    int main() {
        scanf("%d%d", &N, &M);
        memset(num, 0, sizeof(num));
        for(int i = 0; i < N; i ++) {
            scanf("%d", &a[i]);
            long long ans = 0, maxx = 0;
            long long sum = M - a[i];
            for(int j = 1; j <= 100; j ++) {
                long long temp = min(sum / j, num[j]);
                maxx += temp;
                sum -= temp * j;
            }
            num[a[i]] ++;
            printf("%lld%s", i - maxx, i != N - 1 ? " " : "
    ");
        }
        
        return 0;
    }
    View Code

    C2. Exam in BerSU (hard version)

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 2e5 + 10;
    int N, M;
    int a[maxn];
    long long num[110];
    
    int main() {
        scanf("%d%d", &N, &M);
        memset(num, 0, sizeof(num));
        for(int i = 0; i < N; i ++) {
            scanf("%d", &a[i]);
            long long ans = 0, maxx = 0;
            long long sum = M - a[i];
            for(int j = 1; j <= 100; j ++) {
                long long temp = min(sum / j, num[j]);
                maxx += temp;
                sum -= temp * j;
            }
            num[a[i]] ++;
            printf("%lld%s", i - maxx, i != N - 1 ? " " : "
    ");
        }
        
        return 0;
    }
    View Code

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/12238136.html
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