AtCoder Grand Contest 025

最新推荐文章于 2020-05-23 23:25:19 发布

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A - Digits Sum

Time limit : 2sec / Memory limit : 1024MB

Score : 200 points

Problem Statement

Takahashi has two positive integers A and B.

It is known that A plus B equals N. Find the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B" (in base 10).

Constraints

2≤N≤105
N is an integer.

Input

Input is given from Standard Input in the following format:

Output

Print the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B".

Sample Input 1

Copy

Sample Output 1

Copy

When A=2 and B=13, the sums of their digits are 2 and 4, which minimizes the value in question.

Sample Input 2

Copy

Sample Output 2

Copy

#include<bits/stdc++.h>
#define ll long long
using namespace std;
 
int tmp(int n){
    int ans=0;
    while(n){ans+=n%10;n/=10;}
    return ans;
}
 
int main (){
    int n;scanf("%d",&n);
    int ans=0x3f3f3f3f;
    for(int i=1;i<n;i++){
        ans=min(tmp(i)+tmp(n-i),ans);
    }
    printf("%d\n",ans);
    return 0;
}
/**
题意:给定一个数字n 当a+b==n时a的各位数字之和和b的各位数字之和最小;
分析：暴力是直接能过的 
如果要达到最小  显然不进位的加法是满足条件的  就直接是该数的各位数字之和 若该数为10的次方 则最后的ans=10；
int ans=0;
while(n){
   ans+=n%10;
   n/=10;
}
cout<<ans<<endl;
*/

B - RGB Coloring

Time limit : 2sec / Memory limit : 1024MB

Score : 700 points

Problem Statement

Takahashi has a tower which is divided into N layers. Initially, all the layers are uncolored. Takahashi is going to paint some of the layers in red, green or blue to make a beautiful tower. He defines the beauty of the tower as follows:

The beauty of the tower is the sum of the scores of the N layers, where the score of a layer is A if the layer is painted red, A+B if the layer is painted green, B if the layer is painted blue, and 0 if the layer is uncolored.

Here, A and B are positive integer constants given beforehand. Also note that a layer may not be painted in two or more colors.

Takahashi is planning to paint the tower so that the beauty of the tower becomes exactly K. How many such ways are there to paint the tower? Find the count modulo 998244353. Two ways to paint the tower are considered different when there exists a layer that is painted in different colors, or a layer that is painted in some color in one of the ways and not in the other.

Constraints

1≤N≤3×105
1≤A,B≤3×105
0≤K≤18×1010
All values in the input are integers.

Input

Input is given from Standard Input in the following format:

N A B K

Output

Print the number of the ways to paint tiles, modulo 998244353.

Sample Input 1

Copy

4 1 2 5

Sample Output 1

Copy

In this case, a red layer worth 1 points, a green layer worth 3 points and the blue layer worth 2 points. The beauty of the tower is 5 when we have one of the following sets of painted layers:

1 green, 1 blue
1 red, 2 blues
2 reds, 1 green
3 reds, 1 blue

The total number of the ways to produce them is 40.

Sample Input 2

Copy

2 5 6 0

Sample Output 2

Copy

The beauty of the tower is 0 only when all the layers are uncolored. Thus, the answer is 1.

Sample Input 3

Copy

90081 33447 90629 6391049189

Sample Output 3

Copy

577742975

Submit

#include<bits/stdc++.h>
#define ll unsigned long long
using namespace std;

const int mod=998244353;
const int maxn=4e5+7;
ll fac[maxn],inv[maxn];

ll power(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1) ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}

void init(){
    fac[0]=1,fac[1]=1;
    for(int i=2;i<=maxn;i++) fac[i]=fac[i-1]*1ll*i%mod;
    inv[maxn]=power(fac[maxn],mod-2)*1ll;
    for(int i=maxn-1;i>=0;i--) inv[i]=inv[i+1]*1ll*(i+1)%mod;
}

ll c(ll n,ll m){
    if(m==0) return 1;
    if(n<m) return 0;
    if(n==m) return 1;
    return fac[n]*1ll*inv[m]*1ll%mod*inv[n-m]%mod;
}


int main (){
    init();
    int n,a,b;
    ll k,ans=0;
    scanf("%d %d %d %lld",&n,&a,&b,&k);
    if(k==0) { puts("1"); return 0;}
    for(int  x=1;x<=n;x++){
        if((k-a*x)%b) continue;
        ll y=(k-a*x)/b;
        ans=(ans+c(n,x)*1ll%mod*c(n,y)*1ll%mod)%mod;
    }
    printf("%I64d\n",ans);
    return 0;
}
/**
Tricks&Orz.....RE需谨慎！！！！！

题意：输入4个数 n,a,b,k  选的颜色的最多的数量  涂红色所能获得的分 涂蓝色所能获得的分 获得分数之和为k
其实和颜色没有很大的关系;
大意:选择若干个a,b和a+b,使得所选数的总和等于k的方案数;

分析：不要被a+b干扰了  就差不多能出来了  2333...
首先我们可以得到  a*x+b*y=k  枚举x 得到y ans=ans+c(n,x)*c(n,y);
大概意思就是 给定一个初始值为0的数组 然后让你将a和b填入该数组中 只要和为k就NICE了 问方案数 想清楚就可以miao了
 题目质量很良心 RE都能卡我这么久的
 
*/