AtCoder Beginner Contest 104

A - Rated for Me

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $100100$ points

Problem Statement

A programming competition site AtCode regularly holds programming contests.

The next contest on AtCode is called ABC, which is rated for contestants with ratings less than $12001200$.

The contest after the ABC is called ARC, which is rated for contestants with ratings less than $28002800$.

The contest after the ARC is called AGC, which is rated for all contestants.

Takahashi's rating on AtCode is $\mathrm{RR}$. What is the next contest rated for him?

Constraints

• $\mathrm{0\le R\le 42080\le R\le 4208}$
• $\mathrm{RR}$ is an integer.

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Input

Input is given from Standard Input in the following format:

$\mathrm{RR}$

Output

Print the name of the next contest rated for Takahashi (ABC, ARC or AGC).

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Sample Input 1 Copy

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1199

Sample Output 1 Copy

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ABC

$11991199$ is less than $12001200$, so ABC will be rated.

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Sample Input 2 Copy

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1200

Sample Output 2 Copy

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ARC

$12001200$ is not less than $12001200$ and ABC will be unrated, but it is less than $28002800$ and ARC will be rated.

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Sample Input 3 Copy

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4208

Sample Output 3 Copy

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AGC
代码：

import java.util.*;

public class Main {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        if(n < 1200)System.out.println("ABC");
        else if(n < 2800)System.out.println("ARC");
        else System.out.println("AGC");
    }
}

B - AcCepted

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $200200$ points

Problem Statement

You are given a string $\mathrm{SS}$. Each character of $\mathrm{SS}$ is uppercase or lowercase English letter. Determine if $\mathrm{SS}$ satisfies all of the following conditions:

• The initial character of $\mathrm{SS}$ is an uppercase A.
• There is exactly one occurrence of C between the third character from the beginning and the second to last character (inclusive).
• All letters except the A and C mentioned above are lowercase.

Constraints

• $\mathrm{4\le |S|\le 104\le |S|\le 10}$ ($|S||S|$ is the length of the string $\mathrm{SS}$.)
• Each character of $\mathrm{SS}$ is uppercase or lowercase English letter.

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Input

Input is given from Standard Input in the following format:

$\mathrm{SS}$

Output

If $\mathrm{SS}$ satisfies all of the conditions in the problem statement, print AC; otherwise, print WA.

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Sample Input 1 Copy

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AtCoder

Sample Output 1 Copy

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AC

The first letter is A, the third letter is C and the remaining letters are all lowercase, so all the conditions are satisfied.

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Sample Input 2 Copy

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ACoder

Sample Output 2 Copy

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WA

The second letter should not be C.

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Sample Input 3 Copy

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AcycliC

Sample Output 3 Copy

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WA

The last letter should not be C, either.

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Sample Input 4 Copy

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AtCoCo

Sample Output 4 Copy

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WA

There should not be two or more occurrences of C.

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Sample Input 5 Copy

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Atcoder

Sample Output 5 Copy

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WA

The number of C should not be zero, either.

代码：

import java.util.*;

public class Main {
    static boolean check(String s) {
        if(s.charAt(0) != 'A')return false;
        boolean flag = false;
        for(int i = 1;i < s.length();i ++) {
            if(s.charAt(i) != 'C') {
                if(s.charAt(i) >= 'A' && s.charAt(i) <= 'Z')return false;
            }
            else if(flag || i < 2 || s.length() - i < 2)return false;
            else flag = true;
        }
        return flag;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.nextLine();
        if(check(s)) {
            System.out.println("AC");
        }
        else {
            System.out.println("WA");
        }
    }
}

C - All Green

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $300300$ points

Problem Statement

A programming competition site AtCode provides algorithmic problems. Each problem is allocated a score based on its difficulty. Currently, for each integer $\mathrm{ii}$between $11$ and $\mathrm{DD}$ (inclusive), there are ${\mathrm{pipi}}^{}$ problems with a score of $\mathrm{100i100i}$ points. These ${\mathrm{p1+\dots +pDp1+\dots +pD}}^{}$ problems are all of the problems available on AtCode.

A user of AtCode has a value called total score. The total score of a user is the sum of the following two elements:

• Base score: the sum of the scores of all problems solved by the user.
• Perfect bonuses: when a user solves all problems with a score of $\mathrm{100i100i}$ points, he/she earns the perfect bonus of ${\mathrm{cici}}^{}$ points, aside from the base score $(1\le i\le D)(1\le i\le D)$.

Takahashi, who is the new user of AtCode, has not solved any problem. His objective is to have a total score of $\mathrm{GG}$ or more points. At least how many problems does he need to solve for this objective?

Constraints

• $\mathrm{1\le D\le 101\le D\le 10}$
• $\mathrm{1\le pi\le 1001\le pi\le 100}$
• $\mathrm{100\le ci\le 106100\le ci\le 106}$
• $\mathrm{100\le G100\le G}$
• All values in input are integers.
• ${\mathrm{cici}}^{}$ and $\mathrm{GG}$ are all multiples of $100100$.
• It is possible to have a total score of $\mathrm{GG}$ or more points.

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Input

Input is given from Standard Input in the following format:

$\mathrm{DD}$ $\mathrm{GG}$
${\mathrm{p1p1}}^{}$ ${\mathrm{c1c1}}^{}$
$::$
${\mathrm{pDpD}}^{}$ ${\mathrm{cDcD}}^{}$

Output

Print the minimum number of problems that needs to be solved in order to have a total score of $\mathrm{GG}$ or more points. Note that this objective is always achievable (see Constraints).

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Sample Input 1 Copy

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2 700
3 500
5 800

Sample Output 1 Copy

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3

In this case, there are three problems each with $100100$ points and five problems each with $200200$ points. The perfect bonus for solving all the $100100$-point problems is $500500$ points, and the perfect bonus for solving all the $200200$-point problems is $800800$ points. Takahashi's objective is to have a total score of $700700$ points or more.

One way to achieve this objective is to solve four $200200$-point problems and earn a base score of $800800$ points. However, if we solve three $100100$-point problems, we can earn the perfect bonus of $500500$ points in addition to the base score of $300300$ points, for a total score of $800800$ points, and we can achieve the objective with fewer problems.

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Sample Input 2 Copy

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2 2000
3 500
5 800

Sample Output 2 Copy

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7

This case is similar to Sample Input 1, but the Takahashi's objective this time is $20002000$ points or more. In this case, we inevitably need to solve all five $200200$-point problems, and by solving two $100100$-point problems additionally we have the total score of $20002000$ points.

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Sample Input 3 Copy

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2 400
3 500
5 800

Sample Output 3 Copy

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2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is $400400$ points or more. In this case, we only need to solve two $200200$-point problems to achieve the objective.

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Sample Input 4 Copy

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5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4 Copy

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66

There is only one $500500$-point problem, but the perfect bonus can be earned even in such a case.

可以dfs遍历所有的情况，直到大于等于所需要的为止。

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define Max 100001
#define inf 1000000000000000
using namespace std;
struct title {
    int p,c,score;
}s[11];
int d,g,ans = 1000;
int vis[11];
void dfs(int sum,int k) {
    for(int i = 1;i <= d;i ++) {
        if(vis[i])continue;
        if(g - sum <= s[i].score) {
            int temp;
            if(g - sum < s[i].p * i) {
                temp = g - sum;
                temp = temp / i + (temp % i > 0);
            }
            else temp = s[i].p;
            ans = min(ans,k + temp);
            continue;
        }
        vis[i] = 1;
        dfs(sum + s[i].score,k + s[i].p);
        vis[i] = 0;
    }
}
int main() {
    scanf("%d%d",&d,&g);
    g /= 100;
    for(int i = 1;i <= d;i ++) {
        scanf("%d%d",&s[i].p,&s[i].c);
        s[i].c /= 100;
        s[i].score = s[i].p * i + s[i].c;
    }
    dfs(0,0);
    printf("%d",ans);
}

因为最多十种题目，可以用二进制位存状态，遍历各种情况。

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define Max 100001
#define inf 1000000000000000
using namespace std;
struct problem {
    int p,c,score;
}s[11];
int d,g,ans = 1000;

int main() {
    scanf("%d%d",&d,&g);
    g /= 100;
    for(int i = 1;i <= d;i ++) {
        scanf("%d%d",&s[i].p,&s[i].c);
        s[i].c /= 100;
        s[i].score = s[i].p * i + s[i].c;
    }
    for(int i = 1;i < 1 << (d + 1);i ++) {
        int sum = 0,k = 0,maxn;
        for(int j = 1;j <= d;j ++) {
            if(i & (1 << j)) {
                sum += s[j].score;
                k += s[j].p;
            }
            else maxn = j;
        }
        if(sum < g) {
            int need = (g - sum + maxn - 1) / maxn;
            if(need > s[maxn].p)continue;
            k += need;
        }
        ans = min(ans,k);
    }
    printf("%d",ans);
}

D - We Love ABC

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $400400$ points

Problem Statement

The ABC number of a string $\mathrm{TT}$ is the number of triples of integers $(i,j,k)(i,j,k)$ that satisfy all of the following conditions:

• $\mathrm{1\le i<j<k\le |T|1\le i<j<k\le |T|}$ ($|T||T|$ is the length of $\mathrm{TT}$.)
• ${\mathrm{Ti=Ti=}}^{}$ A (${\mathrm{TiTi}}^{}$ is the $\mathrm{ii}$-th character of $\mathrm{TT}$ from the beginning.)
• ${\mathrm{Tj=Tj=}}^{}$ B
• ${\mathrm{Tk=Tk=}}^{}$ C

For example, when $\mathrm{T=T=}$ ABCBC, there are three triples of integers $(i,j,k)(i,j,k)$ that satisfy the conditions: $(1,2,3),(1,2,5),(1,4,5)(1,2,3),(1,2,5),(1,4,5)$. Thus, the ABC number of $\mathrm{TT}$ is $33$.

You are given a string $\mathrm{SS}$. Each character of $\mathrm{SS}$ is A, B, C or ?.

Let $\mathrm{QQ}$ be the number of occurrences of ? in $\mathrm{SS}$. We can make ${\mathrm{3Q3Q}}^{}$ strings by replacing each occurrence of ? in $\mathrm{SS}$ with A, B or C. Find the sum of the ABC numbers of all these strings.

This sum can be extremely large, so print the sum modulo ${\mathrm{109+7109+7}}^{}$.

Constraints

• $\mathrm{3\le |S|\le 1053\le |S|\le 105}$
• Each character of $\mathrm{SS}$ is A, B, C or ?.

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Input

Input is given from Standard Input in the following format:

$\mathrm{SS}$

Output

Print the sum of the ABC numbers of all the ${\mathrm{3Q3Q}}^{}$ strings, modulo ${\mathrm{109+7109+7}}^{}$.

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Sample Input 1 Copy

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A??C

Sample Output 1 Copy

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8

In this case, $\mathrm{Q=2Q=2}$, and we can make ${\mathrm{3Q=93Q=9}}^{}$ strings by by replacing each occurrence of ? with A, B or C. The ABC number of each of these strings is as follows:

• AAAC: $00$
• AABC: $22$
• AACC: $00$
• ABAC: $11$
• ABBC: $22$
• ABCC: $22$
• ACAC: $00$
• ACBC: $11$
• ACCC: $00$

The sum of these is $\mathrm{0+2+0+1+2+2+0+1+0=80+2+0+1+2+2+0+1+0=8}$, so we print $88$ modulo ${\mathrm{109+7109+7}}^{}$, that is, $88$.

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Sample Input 2 Copy

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ABCBC

Sample Output 2 Copy

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3

When $\mathrm{Q=0Q=0}$, we print the ABC number of $\mathrm{SS}$ itself, modulo ${\mathrm{109+7109+7}}^{}$. This string is the same as the one given as an example in the problem statement, and its ABC number is $33$.

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Sample Input 3 Copy

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????C?????B??????A???????

Sample Output 3 Copy

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979596887

In this case, the sum of the ABC numbers of all the ${\mathrm{3Q3Q}}^{}$ strings is $22919796129242291979612924$, and we should print this number modulo ${\mathrm{109+7109+7}}^{}$, that is, $979596887979596887$.

动态规划。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define Max 100005
using namespace std;
typedef long long LL;
const int Mod = 1000000007;
char s[Max];
LL dp[4][Max];///0 '?' 1 'A' 2 'AB' 3 'ABC'
int a,b;
int main() {
    scanf("%s",s + 1);
    dp[0][0] = 1;///初始为1
    int n = strlen(s + 1);
    for(int i = 1;i <= n;i ++) {
        for(int j = 0;j <= 3;j ++) {
            dp[j][i] = dp[j][i - 1];///加上之前的
            if(s[i] == '?')dp[j][i] = (dp[j][i] * 3) % Mod;///如果是问号 有三种选择
            if(j && (s[i] == '?' || s[i] - 'A' + 1 == j)) {///如果对应于匹配位置
                dp[j][i] = (dp[j][i] + dp[j - 1][i - 1]) % Mod;
            }
        }
    }
    printf("%lld",dp[3][n]);
}