2018HPU暑期集训第二次积分赛_场外同步赛

又是斐波那契数列??(HPUOJ 1471)

Description:

大家都知道斐波那契数列吧?斐波那契数列的定义是这样的: f 0 = 0 f_0 = 0 f0=0; f 1 = 1 f_1 = 1 f1=1; f i = f i − 1 + f i − 2 f_i = f_{i-1} + f_{i-2} fi=fi1+fi2 现在给你一个数x,聪明的你一定知道这是斐波那契数列中的第几项。 (数据保证x一定有对应的项y,且 2 <= y < 1e4)

Input:

第一行一个整数T,表示测试组数。
之后的T行,每行一个数x

Output:

对于每个测试数据,输出一行表示数x是第几项

Sample Input:

2
2
5

Sample Output:

3
5

题目链接

递推算出1e4之内的所有斐波那契数列,建立map映射映射到下标即可。

我以为最大数据会很大还写了一个string大数加法斐波那契数列,结果TLE…

2018HPU暑期集训第二次积分赛_场外同步赛_第1张图片

太蠢了…这道题目直接用unsigned long long就可以。

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
	Finish_read = 1;
	x = 0;
	int f = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') {
			f = -1;
		}
		if (ch == EOF) {
			return;
		}
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	x *= f;
	Finish_read = 1;
};

ull f[maxn];
map<ull, int> m;

int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
	f[0] = 0;
	f[1] = 1;
	for (int i = 2; i < maxn; ++i) {
		f[i] = f[i - 1] + f[i - 2];
		m[f[i]] = i;
	}
	int t;
	read(t);
	for (int Case = 1; Case <= t; ++Case) {
		ull x;
		read(x);
		cout << m[x] << endl;
	}
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

Simplest (HPUOJ 1473)

Description:

给你一段由数字0 - 9组成的字符串,请你输出它的最简自然数形式

Input:

第一行一个T,表示T组测试数据.(1 ≤ T ≤ 100)
每一组数据占一行,每一行一串字符 S. (1 ≤ strlen(S) ≤ 100000)

Output:

输出字符串对应的最简自然数形式

Sample Input:

2
2018722
02018722

Sample Output:

2018722
2018722

题目链接

字符串读取,把数字前面的0去掉即可,注意特判0。

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4+5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
	Finish_read = 1;
	x = 0;
	int f = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-')
			f = -1;
		if (ch == EOF)
			return;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	x *= f;
	Finish_read = 1;
};
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
	int t;
	cin >> t;
	while (t--) {
		string str;
		cin >> str;
		bool flag = 0;
		for (int i = 0; i < int(str.length()); ++i) {
			if (str[i] != '0') {
				flag = 1;
			}
			if (flag) {
				cout << str[i];
			}
		}
		if (!flag) {
			cout << 0;
		}
		cout << endl;
	}
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

zz’s math problem Ⅰ (HPUOJ 1474)

Description:

有些人心如花木,皆向阳而生
|《剑来》
zz很喜欢数学,但是他又是一个数学渣,我们称这种生物叫做学渣,
zz又碰到了一个数学小问题,定义一个函数P (x)
例如:P (123) = 1! ∗ 2! ∗ 3! 现在需要找到的就是最大的大于等于x大的数z的函数值和x相等,
即P (z) = P (x) 当然z这个数不能包含1或者0
还请输出最大的符合要求数z(不一定比x大)

Input:

第1行输入T (1 ≤ T ≤ 20)组数据
第2行输入n(1 ≤ n ≤ 100),表示x的数字个数
第3行输入正整数 x

Output:

输出最大的z(数据保证x内含大于1的数,所以z必定有解)

Sample Input:

2
4
1234
3
555

Sample Output:

33222
555

题目链接

对每一个数进行分解,分解为非0、1数字的阶乘相乘,把所有分解的数字按照降序输出即可。

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4 + 5;
const int mod = 1e5 + 3;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
    Finish_read = 1;
    x = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') {
            f = -1;
        }
        if (ch == EOF) {
            return;
        }
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= f;
    Finish_read = 1;
};
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int c[10];
    int t;
    read(t);
    while (t--) {
        mem(c, 0);
        int n;
        read(n);
        char a;
        while (n--) {
            cin >> a;
            if (a == '4') {
                c[2] += 2;
                c[3]++;
            }
            else if (a == '6') {
                c[3]++;
                c[5]++;
            }
            else if (a == '8') {
                c[2] += 3;
                c[7]++;
            }
            else if (a == '9') {
                c[2]++;
                c[3] += 2;
                c[7]++;
            }
            else {
                c[a - '0']++;
            }
        }
        for (int i = 9; i > 1; --i) {
            if (c[i]) {
                while (c[i]--) {
                    cout << i;
                }
            }
        }
        cout << endl;
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

zz’s math problem Ⅱ (HPUOJ 1475)

Description:

zz作为一个数学盲也认为这个数学题真的很简单, 学弟学妹们终于可以顺利签到了qwq
给出 N N N个正整数 a 1 , a 2 , . . . , a N a_1,a_2,...,a_N a1,a2,...,aN
我们寻找一个这个表达式的最大的值 f ( m ) = ( m m o d a 1 ) + ( m m o d a 2 ) + . . . + ( m m o d a N ) f(m)=(m mod a1)+(m mod a2)+...+(m mod aN) f(m)=(mmoda1)+(mmoda2)+...+(mmodaN)

m o d mod mod的意思即为 A / B A/B A/B的余数

Input:

1 1 1行输入 T ( 1 ≤ T ≤ 20 ) T(1≤T≤20) T(1T20)组数据
2 2 2行输入 N ( 1 ≤ N ≤ 1 e 3 ) N(1≤N≤1e3) N(1N1e3)
3 3 3行输入 n n n个数字 a i ( 1 ≤ a i ≤ 1 e 5 ) a_i(1≤ai≤1e5) ai(1ai1e5),

Output:

输出 f f f的最大值

Sample Input:

1
3
3 4 6

Sample Output:

10

题目链接

一定有一个数num使所有 n u m % a i = a i − 1 num\%a_i=a_i-1 num%ai=ai1成立。

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
    Finish_read = 1;
    x = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') {
            f = -1;
        }
        if (ch == EOF) {
            return;
        }
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= f;
    Finish_read = 1;
};
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int t;
    read(t);
    while (t--) {
        int n;
        read(n);
        vector<int> a(n);
        ll ans = 0;
        for (int i = 0; i < n; ++i) {
            read(a[i]);
            ans += a[i] - 1;
        }
        printf("%lld\n", ans);
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

Operation on sequence (HPUOJ 1478)

Description:

井底之蛙,偶见圆月,便欣然忘忧。

\hspace{5cm} —《剑来》

为什么总是要为难学弟学妹呢,zz并不想这样,zz决定把出一道简单的题来福利学弟学妹们。

一组下标从1开始的数组s,进行q次操作:

考虑两种操作:

1 r,将子序列a[1] 到 a[r] 从小到大排序

2 r,将子序列a[1] 到 a[r] 从大到小排序

Input:

第一行输入T组数据 T ( 1 ≤ T ≤ 10 ) T(1≤T≤10) T(1T10)

第一行输入两个数字 n , q ( 1 ≤ n , q ≤ 1 e 4 ) n,q(1≤n,q≤1e4) n,q(1n,q1e4)

第二行包含n个整数 a i ( − 1 e 9 ≤ a i ≤ 1 e 9 ) a_i(−1e9≤a_i≤1e9) ai(1e9ai1e9) — 初始序列

然后q行表示m个操作. 第i行包含两个整数 o p ( 1 ≤ o p ≤ 2 ) , r ( 1 ≤ r ≤ n ) op(1≤op≤2), r(1≤r≤n) op(1op2),r(1rn)

Sample Input:

2
3 1
1 2 3
2 2
4 2
1 2 4 3
2 3
1 2

Sample Output:

2 1 3
2 4 1 3

题目链接

从后往前查找,只进行范围比之前大的操作。

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
    Finish_read = 1;
    x = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') {
            f = -1;
        }
        if (ch == EOF) {
            return;
        }
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= f;
    Finish_read = 1;
};
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int t;
    read(t);
    for (int Case = 1, n, q; Case <= t; ++Case) {
        read(n); read(q);
        vector<int> a(n);
        for (int i = 0; i < n; ++i) {
            read(a[i]);
        }
        vector<PII> operate(q);
        for (int i = 0; i < q; ++i) {
            read(operate[i].first);
            read(operate[i].second);
        }
        stack<PII> op;
        int temp = -1;
        for (int i = q - 1; i >= 0; --i) {
            if (operate[i].second > temp) {
                op.push(operate[i]);
                temp = operate[i].second;
            }
        }
        while (!op.empty()) {
            sort(a.begin(), a.begin() + op.top().second);
            if (op.top().first == 2) {
                reverse(a.begin(), a.begin() + op.top().second);
            }
            op.pop();
        }
        for (auto i : a) {
            printf("%d ", i);
        }
        printf("\n");
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

Operation on sequence (HPUOJ 1477)

Description:

真·签到题
输入一个表达式A op B op操作为′+′,′−′,′∗′ ,A,B为整数

Input:

第一行输入T 组数据 T ( 1 ≤ T ≤ 1 e 2 ) T(1≤T≤1e2) T(1T1e2)
第二行输入 A o p B ( − 1 e 9 ≤ A , B ≤ 1 e 9 ) A op B (−1e9≤A,B≤1e9) AopB(1e9A,B1e9) — 初始序列

Output:

输出表达式结果并换行即可

Sample Input:

1
1 + 1

Sample Output:

2

题目链接

emm…

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4 + 5;
const int mod = 1e5 + 3;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
	Finish_read = 1;
	x = 0;
	int f = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') {
			f = -1;
		}
		if (ch == EOF) {
			return;
		}
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	x *= f;
	Finish_read = 1;
};
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
	int t;
	cin >> t;
	while (t--) {
		ll a, b, ans;
		char op;
		cin >> a >> op >> b;
		if (op == '+') {
			ans = a + b;
		}
		else if (op == '-') {
			ans = a - b;
		}
		else {
			ans = a * b;
		}
		cout << ans << endl;
	}
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

又是划分问题 (HPUOJ 1480)

Description:

给你一个正整数n,将其划分,要求划分成的数必须是2的幂,有多少种划分方法??
结果可能很大,我们输出对1e9+7取模的结果

Input:

一个正整数n,代表要划分的数;
1 < = n < = 1 e 7 1<=n<=1e7 1<=n<=1e7

Output:

输出可划分的方法数

Sample Input:

15
67

Sample Output:

26
2030

Hint:

当n=6时,我们可以将其划分为
1 1 1 1 1 1
1 1 1 1 2
1 1 2 2
2 2 2
1 1 4
2 4
这6种划分方法

题目链接

和这道题一样

POJ 2229 HDU 2709 Sumsets

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e7 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
	Finish_read = 1;
	x = 0;
	int f = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') {
			f = -1;
		}
		if (ch == EOF) {
			return;
		}
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	x *= f;
	Finish_read = 1;
};
 
ll dp[maxn];
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
	int n;
	while (~scanf("%d", &n)) {
		dp[0] = dp[1] = 1;
		for (int i = 2; i <= n; ++i) {
			if (i & 1) {
				dp[i] = dp[i - 1];
				dp[i] %= mod;
			}
			else {
				dp[i] = dp[i - 2] + dp[i / 2];
				dp[i] %= mod;
			}
		}
		printf("%lld\n", dp[n]);
	}
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

台阶问题 (HPUOJ 1479)

Description:

有 N 级的台阶,你一开始在底部,每次可以向上迈最多 K 级台阶(最少 1 级),问到达第 N 级台阶有多少种不同方式。

Input:

多组输入,两个正整数N(N ≤ 1000),K(K ≤ 100)。

Output:

一个正整数,为不同方式数,由于答案可能很大,你需要输出 ans mod 100003 后的结果。

Sample Input:

5 2

Sample Output:

8

题目链接

每次走 n n n阶楼梯可以选择先走一格,再走剩下 n − 1 n-1 n1格即为 f [ n − 1 ] f[n-1] f[n1],或者先走 2 2 2格,再走剩下 n − 2 n-2 n2格即为 f [ n − 2 ] f[n-2] f[n2],以此类推,把每种走法都加起来即为结果。

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4 + 5;
const int mod = 1e5 + 3;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
    Finish_read = 1;
    x = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') {
            f = -1;
        }
        if (ch == EOF) {
            return;
        }
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= f;
    Finish_read = 1;
};
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int n, k;
    while (~scanf("%d %d", &n, &k)) {
        vector<ll> f(n + 1, 0);
        f[0] = 1;
        f[1] = 1;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                if (i - j >= 0) {
                    f[i] += f[i - j];
                    f[i] %= mod;
                }
            }
        }
        printf("%lld\n", f[n]);
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

括号括号 (HPUOJ 1476)

Description:

小明今年上大学,在大学里发现有很多同学都女朋友,两人整天都在一起腻歪,小明看到后感觉很孤单,现在,给你一行括号序列,你来判断一下其中的括号是否配对。

Input:

多组输入,每一组第一行输入一个数T(0<

Output:

每组输入数据的输出占一行,如果该字符串中所含的括号是配对的,则输出Yes,如果不配对则输出No。

Sample Input:

3
[(])
(])
([])

Sample Output:

No
No
Yes

题目链接

栈的入门题目括号匹配。

AC代码:

#include 
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
const int INF = 0x3f3f3f40;
const int maxn = 1e4 + 5;
const int mod = 1e5 + 3;
const double eps = 2e-8;
const double pi = asin(2.0) * 2;
const double e = 3.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
    Finish_read = 1;
    x = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') {
            f = -1;
        }
        if (ch == EOF) {
            return;
        }
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= f;
    Finish_read = 1;
};
 
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int n;
    while (cin >> n) {
        string str;
        bool flag;
        while (n--) {
            cin >> str;
            flag = 1;
            stack<char> a;
            for (int i = 0; i < int(str.length()); ++i) {
                if (str[i] == '(' || str[i] == '[') {
                    a.push(str[i]);
                }
                else {
                    if (a.empty()) {
                        flag = 0;
                        break;
                    }
                    else {
                        if (str[i] == ')') {
                            if (a.top() == '(') {
                                a.pop();
                            }
                            else {
                                flag = 0;
                                break;
                            }
                        }
                        else {
                            if (str[i] == ']') {
                                if (a.top() == '[') {
                                    a.pop();
                                }
                                else {
                                    flag = 0;
                                    break;
                                }
                            }
                        }
                    }
                }
            }
            if (flag && a.empty()) {
                printf("Yes\n");
            }
            else {
                printf("No\n");
            }
        }
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}

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