codeforces - 1256 (div3)-优快云博客

本文深入解析了一场Codeforces Div.3比赛的四道题目，包括PaymentWithoutChange、MinimizethePermutation、PlatformsJumping和BinaryStringMinimizing。通过详细展示每题的解题思路及代码实现，特别是对贪心算法的应用，为读者提供了宝贵的算法竞赛经验。

令我感到非常自闭的一场div3

反正没打完就溜了

A - Payment Without Change

思维

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#pragma GCC optimize(2)
 
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second
 
using namespace std;
typedef long long ll;
typedef pair<int, int > PII;
 
const int N = 1e6 + 5, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int q;
ll a, b, n, s;
ll tmp1, tmp2;
 
int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//    cin.tie(0);
//    cout.tie(0);
//    ios::sync_with_stdio(0);
    cin >> q;
    while (q--) {
        cin >> a >> b >> n >> s;
        tmp1 = s / n;
        tmp2 = s % n;
        if (tmp1 <= a) {
            if (b >= tmp2)    puts("YES");
            else puts("NO");
        } else {
            if (a * n + b >= s)    puts("YES");
            else    puts("NO");
        }
    }
}

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B - Minimize the Permutation

贪心，具体怎么贪忘了，前段时间写的

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#pragma GCC optimize(2)
 
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second
 
using namespace std;
typedef long long ll;
typedef pair<int, int > PII;
 
const int N = 110, mod = 1e9 + 9, INF = 0x3f3f3f3f;
int q, n, idx, sum, pos;
int a[N];
bool vis[N];
 
void Get() {
    idx = INF;
    for (int i = 1; i <= n; ++i) {
        if (vis[i]) continue;
        if (a[i] < idx) {
            idx = a[i];
            pos = i;
        }
    }
    return ;
}
 
int main()
{
    cin >> q;
    while (q--) {
        mm(vis, 0);
        cin >> n;
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
        }
        
        int i, j;
        sum = 0;
        while (sum < n) {
            Get();
            printf("%d ", idx);
            vis[pos] = 1;
            sum ++;
            while (1) {
                pos --;
                if (vis[pos] == 0 || pos < 1) break;
            }
            while (1) {
                pos --;
                if (vis[pos] == 0 || pos < 1) break;
            }
//            cout << "pos :" << pos << endl;
            for (j = 1; j <= pos; ++j) {
                if (!vis[j]) {
                    printf("%d ", a[j]);
                    vis[j] = 1;
                    sum ++;
                }
            }
        }
        printf("\n");
    }
}

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C - Platforms Jumping

也是一道贪心，但是没写出来，WA来WA去的WA到怀疑人生

贴一个当时网上找的代码

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 11;
int c[N],ans[N];
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n,m,d;
    cin>>n>>m>>d;//n是宽度，m是板子数目，d是可以最多跳多远
    int pos=0;
    for (int i=1; i<=m; i++) {
        pos+=d;//每次跳最远
        int l;
        cin>>l;
        pos+=l-1;
        c[i]=l;//记录板子长度
    }
    pos+=d;//最后再跳一次
    if (pos<n+1) {//如果跳不过去
        cout<<"NO";//就输出NO
        return 0;
    }
    cout<<"YES\n";
    pos=0;
    int sum=0;
    for (int i=1; i<=m; i++)
        sum+=c[i];//板子总长度
    for (int i=1; i<=m; i++) {
        if (pos+d+sum-1<=n) pos+=d;
        else pos=n-sum+1;
        for (int j=pos; j<=pos+c[i]-1; j++)
            ans[j]=i;
        pos+=c[i]-1;
        sum-=c[i];
    }
    for (int i=1; i<=n; i++)
        cout<<ans[i]<<" ";
}

View Code

D - Binary String Minimizing

感觉这场貌似反而D比较简单？

还是贪心，就尽量把0往前面挪

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <queue>
#pragma GCC optimize(2)
 
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define one first
#define two second
 
using namespace std;
typedef long long ll;
typedef pair<int, int > PII;
 
const int N = 1e6 + 5, mod = 1e9 + 9, INF = 0x3f3f3f3f;
ll n, q, k, ans, idx;
char s[N];
queue <ll> ss;
 
int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
//    cin.tie(0);
//    cout.tie(0);
//    ios::sync_with_stdio(0);
    cin >> q;
    while (q--) {
        while (!ss.empty())
            ss.pop();
        cin >> n >> k;
        scanf("%s", s + 1);
        for (int i = 1; i <= n; ++i) {
            if (s[i] == '0') {
                ss.push(i);
            }
        }
        ans = 0, idx = 1;
        while (!ss.empty()) {
            int x = ss.front();
            ss.pop();
            if (ans + x - idx > k) {
                swap(s[x], s[x + ans - k]);
                break;
            }
            ans += (x - idx);
            swap(s[idx++], s[x]);
        }
        printf("%s\n", s + 1);
    }
}
/*
1
7 9
1111100
*/

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