18年11月5日 NOIP模拟赛

T1

题解

对于k=100的情况，贪心

对于100%的数据

可以发现，当前的决策只对后面的开采有影响，且剩余耐久度与之后的开采收益成正比，如果倒着考虑这个问题，得出i-n的星球1点耐久度所能获得的最大收益，从后往前dp，得出最大值最后乘w就是答案

代码

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100001;
int n,w,t[maxn],a[maxn];
double k,c,ans;
int main()
{
    //freopen("exploit.in","r",stdin);
    //freopen("exploit.out","w",stdout);
    scanf("%d%lf%lf%d",&n,&k,&c,&w);
    k=1-0.01*k;c=1+0.01*c;
    for(int i=1;i<=n;i++)scanf("%d%d",&t[i],&a[i]);
    for(int i=n;i;i--)
        if(t[i]==1)
            ans=max(ans,ans*k+a[i]);
        else
            ans=max(ans,ans*c-a[i]);
    printf("%.2lf
",ans*w);
}

T2

题解

从前往后推出每个人最少/最多有几个和第一个人相同的勋章

然后看最后一个最少是否是0即可

代码

#include<bits/stdc++.h>

inline int read()
{
    int x = 0,t = 1; char ch = getchar();
    while(ch < '0' || ch > '9'){if(ch == '-') t = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = x*10+ch-'0'; ch = getchar();}
    return x*t;
}

const int MAXN = 20010;
int n;
int num[MAXN];
int L,R,mid;
int dp[MAXN],gun[MAXN];

inline void init()
{
    n = read();
    for(int i=1;i<=n;i++)
        num[i] = read();
    for(int i=1;i<n;i++)
        L=std::max(L,num[i]+num[i+1]);
    
    L=std::max(L,num[n]+num[1]),R=0x7ffffff;
}

inline void DP()
{
    while(L<R)
    {
        mid=(L+R)>>1;
        dp[1]=gun[1]=num[1];
        
        for(int i=2;i<=n;i++)
            dp[i]=std::min(num[i],num[1]-gun[i-1]),
        
        gun[i]=std::max(0,num[i]-(mid-num[i-1]-(num[1]-dp[i-1])));
        
        if(gun[n]==0) R=mid;
        else L=mid+1;
    }
    std::cout << L;
}

int main(void)
{
    std::ios_base::sync_with_stdio(false); 

    init();
    DP();

    return 0;
}

T3

题解

30%: O(n ^ 2 * m)暴力判断。

100%: 很显然答案的可能性最多只有n 种，所以我们将所有人的答案按字典序排序后枚举

将每个人的答案作为正确答案来进行判断。由于是判断题，若当前人的答案为正确答
案则零分者的答案也就确定了，那么只需统计出这两种答案的人数判断是否满足题意
即可。这一步使用字符串哈希即可解决。

另外要注意p = 0 和p = q = 0 的情况。

代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

const int N = 3e4 + 2, M = 5e2 + 2, sed = 31, SED = 131, mod = 70177, MOD = 92311;
int n, m, p, q, ans, hash[N], HASH[N];
int top, info[mod], nxt[N * 2], fet[N * 2], cnt[N * 2];
struct node {
    char s[M];
    inline bool operator < (const node &b) const {
        return strcmp(s, b.s) < 0;
    }
} a[N];

inline void Insert(const int &x, const int &y) {
    for (int k = info[x]; k; k = nxt[k])
        if (fet[k] == y) {
            ++cnt[k]; return ;
        }
    nxt[++top] = info[x]; info[x] = top;
    fet[top] = y; cnt[top] = 1;
    return ;
}

inline int Query(const int &x, const int &y) {
    for (int k = info[x]; k; k = nxt[k])
        if (fet[k] == y) return cnt[k];
    return 0;
}

inline void Solve1() {
    int tmp, TMP; ans = -1;
    for (int i = 0; i < n; ++i) {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == 'N')) % mod;
            TMP = (TMP * SED + (a[i].s[j] == 'N')) % MOD;
        }
        hash[i] = tmp, HASH[i] = TMP;
        Insert(tmp, TMP);
    }
    for (int i = 0; i < n; ++i)
        if (Query(hash[i], HASH[i]) == p) {
            tmp = TMP = 0;
            for (int j = 0; j < m; ++j) {
                tmp = (tmp * sed + (a[i].s[j] == 'Y')) % mod;
                TMP = (TMP * SED + (a[i].s[j] == 'Y')) % MOD;
            }
            if (Query(tmp, TMP) == q) {
                ans = i; break;
            }
        }
    if (ans != -1) printf("%s
", a[ans].s);
    else 	puts("-1");
    return ;
}

char cur[M];
inline void Solve2() {
    int tmp, TMP; ans = -1;
    for (int i = 0; i < n; ++i) {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == 'N')) % mod;
            TMP = (TMP * SED + (a[i].s[j] == 'N')) % MOD;
        }
        hash[i] = tmp, HASH[i] = TMP;
        Insert(tmp, TMP);
    }
    for (int i = n - 1; i >= 0; --i)
        if (Query(hash[i], HASH[i]) == q) {
            tmp = TMP = 0;
            for (int j = 0; j < m; ++j) {
                tmp = (tmp * sed + (a[i].s[j] == 'Y')) % mod;
                TMP = (TMP * SED + (a[i].s[j] == 'Y')) % MOD;
            }
            if (Query(tmp, TMP) == p) {
                ans = i; break;
            }
        }
    if (ans != -1) {
        for (int i = 0; i < m; ++i)
            cur[i] = a[ans].s[i] == 'N' ? 'Y' : 'N';
        printf("%s
", cur);
    }
    else 	puts("-1");
    return ;
}

void Solve3() {
    int tmp, TMP;
    for (int i = 0; i < n; ++i) {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == 'N')) % mod;
            TMP = (TMP * SED + (a[i].s[j] == 'N')) % MOD;
        }
        Insert(tmp, TMP);
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == 'Y')) % mod;
            TMP = (TMP * SED + (a[i].s[j] == 'Y')) % MOD;
        }
        Insert(tmp, TMP);
    }
    bool flag = true;
    for (int i = 0; i < m; ++i) cur[i] = 'N';
    do {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (cur[j] == 'N')) % mod;
            TMP = (TMP * SED + (cur[j] == 'N')) % MOD;
        }
        if (Query(tmp, TMP) == 0) {
            flag = true; break;
        }
        flag = false;
        for (int j = m - 1; j >= 0; --j)
            if (cur[j] == 'Y') cur[j] = 'N';
            else {
                cur[j] = 'Y'; flag = true; break;
            }
    } while (flag);
    if (flag) printf("%s
", cur);
    else 	puts("-1");
    return ;
}

int main() {
    //freopen("answer.in", "r", stdin);
    //freopen("answer.out", "w", stdout);
    scanf("%d%d%d%d", &n, &m, &p, &q);
    for (int i = 0; i < n; ++i) scanf("%s", a[i].s);
    sort(a, a + n);
    if (p) Solve1();
    else if (q) Solve2();
    else 	Solve3();
    fclose(stdin); fclose(stdout);
    return 0;
}