A - Garden

Time limit : 2sec / Memory limit : 1000MB

Score: 100 points

Problem Statement

There is a farm whose length and width are A yard and B yard, respectively. A farmer, John, made a vertical road and a horizontal road inside the farm from one border to another, as shown below: (The gray part represents the roads.)

What is the area of this yard excluding the roads? Find it.

Note

It can be proved that the positions of the roads do not affect the area.

Constraints

  • A is an integer between 2 and 100 (inclusive).
  • B is an integer between 2 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

A B

Output

Print the area of this yard excluding the roads (in square yards).

Sample Input 1

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2 2

Sample Output 1

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1

In this case, the area is 1 square yard.

Sample Input 2

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5 7

Sample Output 2

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24

In this case, the area is 24 square yards.

容斥定理。

代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int b = in.nextInt();
        System.out.println(a * b - a - b + 1);
    }
}

B - 105

Time limit : 2sec / Memory limit : 1000MB

Score: 200 points

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors. Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

  • N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the count.

Sample Input 1

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105

Sample Output 1

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1

Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

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7

Sample Output 2

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0

1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

代码：

import java.util.*;

public class Main {
    static int get(int a) {
        int b = 2;
        for(int i = 3;i <= a / 3;i += 2) {
            if(a % i == 0) {
                b ++;
            }
        }
        return b;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int m = 0;
        if(a >= 105)m ++;
        for(int i = 107;i <= a;i += 2) {
            if(get(i) == 8)m ++;
        }
        System.out.println(m);
    }
}

C - To Infinity

Time limit : 2sec / Memory limit : 1000MB

Score: 300 points

Problem Statement

Mr. Infinity has a string S consisting of digits from 1 to 9. Each time the date changes, this string changes as follows:

  • Each occurrence of 2 in S is replaced with 22. Similarly, each 3 becomes 333, 4 becomes 4444, 5 becomes 55555, 6 becomes 666666, 7 becomes 7777777, 8 becomes 88888888 and 9 becomes 999999999. 1 remains as 1.

For example, if S is 1324, it becomes 1333224444 the next day, and it becomes 133333333322224444444444444444 the day after next. You are interested in what the string looks like after 5×1015 days. What is the K-th character from the left in the string after 5×1015 days?

Constraints

  • S is a string of length between 1 and 100 (inclusive).
  • K is an integer between 1 and 1018 (inclusive).
  • The length of the string after 5×1015 days is at least K.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the K-th character from the left in Mr. Infinity's string after 5×1015 days.

Sample Input 1

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1214
4

Sample Output 1

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2

The string S changes as follows:

  • Now: 1214
  • After one day: 12214444
  • After two days: 1222214444444444444444
  • After three days: 12222222214444444444444444444444444444444444444444444444444444444444444444

The first five characters in the string after 5×1015 days is 12222. As K=4, we should print the fourth character, 2.

Sample Input 2

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3
157

Sample Output 2

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3

The initial string is 3. The string after 5×1015 days consists only of 3.

Sample Input 3

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299792458
9460730472580800

Sample Output 3

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2
只需要看看开头有几个连续的1即可。
代码：
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.nextLine();
        long a = in.nextLong();
        int b = 0;
        for(int i = 0;i < s.length();i ++) {
            if(s.charAt(i) == '1')b ++;
            else break;
        }
        if(b >= a)System.out.println(1);
        else System.out.println(s.charAt(b));
    }
}

 

D - AtCoder Express 2

Time limit : 3sec / Memory limit : 1000MB

Score: 400 points

Problem Statement

In Takahashi Kingdom, there is a east-west railroad and N cities along it, numbered 1, 2, 3, ..., N from west to east. A company called AtCoder Expresspossesses M trains, and the train i runs from City Li to City Ri (it is possible that Li=Ri). Takahashi the king is interested in the following Q matters:

  • The number of the trains that runs strictly within the section from City pi to City qi, that is, the number of trains j such that pi≤Lj and Rj≤qi.

Although he is genius, this is too much data to process by himself. Find the answer for each of these Q queries to help him.

Constraints

  • N is an integer between 1 and 500 (inclusive).
  • M is an integer between 1 and 200 000 (inclusive).
  • Q is an integer between 1 and 100 000 (inclusive).
  • 1≤Li≤Ri≤N (1≤i≤M)
  • 1≤pi≤qi≤N (1≤i≤Q)

Input

Input is given from Standard Input in the following format:

N M Q
L1 R1
L2 R2
:
LM RM
p1 q1
p2 q2
:
pQ qQ

Output

Print Q lines. The i-th line should contain the number of the trains that runs strictly within the section from City pi to City qi.

Sample Input 1

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2 3 1
1 1
1 2
2 2
1 2

Sample Output 1

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3

As all the trains runs within the section from City 1 to City 2, the answer to the only query is 3.

Sample Input 2

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10 3 2
1 5
2 8
7 10
1 7
3 10

Sample Output 2

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1
1

The first query is on the section from City 1 to 7. There is only one train that runs strictly within that section: Train 1. The second query is on the section from City 3 to 10. There is only one train that runs strictly within that section: Train 3.

Sample Input 3

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10 10 10
1 6
2 9
4 5
4 7
4 7
5 8
6 6
6 7
7 9
10 10
1 8
1 9
1 10
2 8
2 9
2 10
3 8
3 9
3 10
1 10

Sample Output 3

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7
9
10
6
8
9
6
7
8
10
简单的区间dp。
代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,q;
int dp[502][502];
int main() {
    int a,b;
    scanf("%d%d%d",&n,&m,&q);
    for(int i = 0;i < m;i ++) {
        scanf("%d%d",&a,&b);
        dp[a][b] ++;
    }
    for(int j = 1;j < n;j ++) {
        for(int i = 1;i + j <= n;i ++) {
            int k = i + j;
            dp[i][k] += dp[i + 1][k] + dp[i][k - 1];
            if(j > 1)dp[i][k] -= dp[i + 1][k - 1];
        }
    }
    for(int i = 0;i < q;i ++) {
        scanf("%d%d",&a,&b);
        printf("%d\n",dp[a][b]);
    }
    return 0;
}

 树状数组，区间向两侧更新，最后查询两侧向内查询。

代码：

#include <iostream>
#include <cstdio>
using namespace std;
int n;
int sum[501][501];
int lowbit(int x) {
    return x&-x;
}
void update(int x,int y) {
    for(int i = x;i > 0;i -= lowbit(i)) {
        for(int j = y;j <= n;j += lowbit(j)) {
            sum[i][j] ++;
        }
    }
}
int get(int x,int y) {
    int ans = 0;
    for(int i = x;i <= y;i += lowbit(i)) {
        for(int j = y;j >= x;j -= lowbit(j)) {
            ans += sum[i][j];
        }
    }
    return ans;
}
int main() {
    int m,q;
    scanf("%d%d%d",&n,&m,&q);
    int a,b;
    for(int i = 0;i < m;i ++) {
        scanf("%d%d",&a,&b);
        update(a,b);
    }
    for(int i = 0;i < q;i ++) {
        scanf("%d%d",&a,&b);
        printf("%d\n",get(a,b));
    }
}

 也可以用邻接表加二分遍历，不过就比较麻烦了。