• CodeForces Round #552 Div.3


    A. Restoring Three Numbers

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int n[5];
    int a, b, c;
    
    int main() {
        for(int i = 0; i < 4; i ++)
            scanf("%d", &n[i]);
        sort(n, n + 4);
        a = n[3] - n[0];
        b = n[3] - n[1];
        c = n[3] - n[2];
        printf("%d %d %d
    ", a, b, c);
        return 0;
    }
    View Code

    B. Make Them Equal

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1010;
    int N;
    int a[maxn], vis[maxn];
    vector<int> ans;
    
    int main() {
        scanf("%d", &N);
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= N; i ++) {
            scanf("%d", &a[i]);
            if(!vis[a[i]]) {
                ans.push_back(a[i]);
                vis[a[i]] = 1;
            }
        }
    
        sort(ans.begin(), ans.end());
        if(ans.size() == 1) printf("0
    ");
        else if(ans.size() == 2) {
            if((ans[1] - ans[0]) % 2)
                printf("%d
    ", ans[1] - ans[0]);
            else printf("%d
    ", (ans[1] - ans[0]) / 2);
        } else if(ans.size() == 3) {
            if(ans[1] - ans[0] == ans[2] - ans[1])
                printf("%d
    ", ans[1] - ans[0]);
            else printf("-1
    ");
        } else printf("-1
    ");
    
        return 0;
    }
    View Code

    C. Gourmet Cat

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int day[20] = {0, 1, 2, 0, 2, 1, 0, 0, 1, 2, 0, 2, 1, 0};
    int a, b, c;
    
    int main() {
        scanf("%d%d%d", &a, &b, &c);
        int ans = 0;
        int t = min(a / 3, min(b / 2, c / 2));
        ans += t * 7;
        a -= (t * 3), b -= (t * 2), c -= (t * 2);
        int na = a, nb = b, nc = c;
        int maxx = 0, temp;
        for(int i = 0; i < 7; i ++) {
            int rec = 0;
            a = na, b = nb, c = nc;
            for(int j = i; j < i + 7; j ++) {
                if(day[j] == 0) {
                    if(a) a --, rec ++;
                    else {
                        maxx = max(maxx, rec);
    
                        break;
                    }
                }
                else if(day[j] == 1) {
                    if(b) b --, rec ++;
                    else {
                        maxx = max(maxx, rec);
                        break;
                    }
                } else {
                    if(c) c --, rec ++;
                    else {
                        maxx = max(maxx, rec);
                        break;
                    }
                }
            }
            //printf("!%d %d   %d
    ", i, maxx, rec);
        }
    
        printf("%d
    ", ans + maxx);
        return 0;
    }
    
    
    /*
    
    700000000 700000000 700000000
    
    */
    View Code

    D. Walking Robot

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 2e5 + 10;
    int N, b, a, na, nb;
    int num[maxn];
    
    int main() {
        scanf("%d%d%d", &N, &b, &a);
        for(int i = 1; i <= N; i ++)
            scanf("%d", &num[i]);
    
        nb = b, na = a;
        int ans;
        for(int i = 1; i <= N; i ++) {
            if(num[i] == 1) {
                if(nb > 0) {
                    if(na + 1 <= a) {
                        na += 1;
                        nb -= 1;
                    } else na -= 1;
                } else na -= 1;
            } else {
                if(na > 0) na -= 1;
                else if(nb > 0) nb -= 1;
            }
            if(na == 0 && nb == 0) {
                ans = i;
                break;
            }
    
            //printf("%d %d
    ", b, a);
        }
        printf("%d
    ", min(ans, N));
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10724136.html
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