AtCoderBeginnerContest104

  • Z时代
  • 2024-01-10
  • 分类：综合

coding

A - Rated for Me

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 RR. What is the next contest rated for him?

Constraints

  • 0≤R≤42080≤R≤4208

  • RR is an integer.

Input

Input is given from Standard Input in the following format:

RR

Output

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

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.

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.

Sample Input 3 Copy

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4208

Sample Output 3 Copy

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AGC
代码：

import java.util.*;
publicclass Main {
publicstaticvoid main(String[] args) {
        Scanner in = new Scanner(System.in);
int n = in.nextInt();
if(n < 1200)System.out.println("ABC");
elseif(n < 2800)System.out.println("ARC");
else System.out.println("AGC");
    }
}

B - AcCepted

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 200200 points

Problem Statement

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

  • The initial character of 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

  • 4≤|S|≤104≤|S|≤10 (|S||S| is the length of the string SS.)

  • Each character of SS is uppercase or lowercase English letter.

Input

Input is given from Standard Input in the following format:

SS

Output

If SS satisfies all of the conditions in the problem statement, print AC; otherwise, print WA.

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.

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.

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.

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.

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.*;
publicclass Main {
staticboolean check(String s) {
if(s.charAt(0) != 'A')returnfalse;
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')returnfalse;
            }
elseif(flag || i < 2 || s.length() - i < 2)returnfalse;
else flag = true;
        }
return flag;
    }
publicstaticvoid 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

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 iibetween 11 and DD (inclusive), there are pipi problems with a score of 100i100i points. These p1+…+pDp1+…+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 100i100i points, he/she earns the perfect bonus of cici points, aside from the base score (1≤i≤D)(1≤i≤D).

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

Constraints

  • 1≤D≤101≤D≤10

  • 1≤pi≤1001≤pi≤100

  • 100≤ci≤106100≤ci≤106

  • 100≤G100≤G

  • All values in input are integers.

  • cici and GG are all multiples of 100100.

  • It is possible to have a total score of GG or more points.

Input

Input is given from Standard Input in the following format:

DDGG
p1p1c1c1
::
pDpDcDcD

Output

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

Sample Input 1 Copy

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2 700
3 500
5 800

Sample Output 1 Copy

Copy

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.

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.

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.

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
usingnamespace 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
usingnamespace 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

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400400 points

Problem Statement

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

  • 1≤i<j<k≤|T|1≤i<j<k≤|T| (|T||T| is the length of TT.)

  • Ti=Ti= A (TiTi is the ii-th character of TT from the beginning.)

  • Tj=Tj= B

  • Tk=Tk= C

For example, when 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 TT is 33.

You are given a string SS. Each character of SS is A, B, C or ?.

Let QQ be the number of occurrences of ? in SS. We can make 3Q3Q strings by replacing each occurrence of ? in 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 109+7109+7.

Constraints

  • 3≤|S|≤1053≤|S|≤105

  • Each character of SS is A, B, C or ?.

Input

Input is given from Standard Input in the following format:

SS

Output

Print the sum of the ABC numbers of all the 3Q3Q strings, modulo 109+7109+7.

Sample Input 1 Copy

Copy

A??C

Sample Output 1 Copy

Copy

8

In this case, Q=2Q=2, and we can make 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 0+2+0+1+2+2+0+1+0=80+2+0+1+2+2+0+1+0=8, so we print 88 modulo 109+7109+7, that is, 88.

Sample Input 2 Copy

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ABCBC

Sample Output 2 Copy

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3

When Q=0Q=0, we print the ABC number of SS itself, modulo 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.

Sample Input 3 Copy

Copy

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

Sample Output 3 Copy

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979596887

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

动态规划。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define Max 100005
usingnamespace std;
typedef longlong LL;
constint 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]);
}

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