51nod做题笔记

2652 阶乘0的数量 V2

因子中2的密度远远大于5的密度，故只需关注因子为5的出现即可，即间隔5，阶乘增加一个零。然后二分答案即可。

1094 和为k的连续区间

这题有$nlogn$的做法，求出前缀和后必有$sum[r]-sum[l-1]=k$，枚举一个$sum[r]$，只需要去找$sum[l-1] + k =sum[r]$即可。map或者二分随便搞。

2462 铺设道路

这题有线性的做法：$a_i > a_{i-1},ans+=a_i-a_{i-1}$.

1506 最小字典序

丑死了

### #include<iostream>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<cstring>
#include<stack>
#include <cmath>

#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
#define inf 0x3f3f3f3f
typedef long long ll;
typedef long double ld;
typedef double db;
typedef std::pair<int, int> pii;
typedef std::pair<ll, ll> pll;
const ll mod = 1e9 + 7;
const int N = 5e5 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;

ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }

inline ll read() {
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

int n;
string s;
int pre[N], suf[N];

int main() {
    io_opt;
    cin >> s;
    n = s.size();
    pre[0] = 0;

    for (int i = 1; i < n; i++) {
        pre[i] = pre[i - 1];
        if (s[i] > s[pre[i - 1]]) {
            pre[i] = i;
        }
    }
    suf[n - 1] = n - 1;
    for (int i = n - 2; i >= 0; i--) {
        suf[i] = suf[i + 1];
        if (s[i] < s[suf[i + 1]]) {
            suf[i] = i;
        }
    }
    if(n<26) {
        string ans = "{";
        for (int i = 1; i < n; i++) {
            swap(s[pre[i - 1]], s[suf[i]]);
            ans = min(ans, s);
            swap(s[pre[i - 1]], s[suf[i]]);
        }
        cout << ans;
    }else{
        for (int i = 0; i < n; i++)
            if (s[i] > s[suf[i]]) {
                swap(s[i], s[suf[i]]);
                break;
            }
        cout << s;
    }
    return 0;
}