每日两小时编程之旅（五）

牛客篇-牛客算法周练9

A-符合条件的数

链接：https://ac.nowcoder.com/acm/contest/5902/A
来源：牛客网

众所周知，某个被共青团点名的学校的机房里有一个蒟蒻，名字叫做clccle，因为文化课成绩不好而经常被班主任gank,这次他遇到了一个很困（rui）难（zhi）的数学题，因为clccle非常辣鸡，所以他想到了聪明的你，你能帮Ta解决这个问题吗？

给定两个整数N,M,表示区间 [2N,2M），请求出在这个区间里有多少个整数i满足i % 7=1

输入描述:

一行，两个整数N,M 0<=N<M<=65

输出描述:

一个数字ans，表示满足条件的数字的个数

示例1

输入

复制

2 3

输出

复制

说明

在区间[22,23 ）里满足条件的整数的个数为零  ，（4，5，6，7，8因为在开区间边界上所以不做考虑）

#include<bits/stdc++.h>

#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
ll n,m;

int main()
{
    cin>>n>>m;
    n=1LL<<n,m=1LL<<m;//在c++中，LL是long long int的意思
    //1LL一般与*共用，即1LL*或*1LL在这种情况下，被赋值的变量将会变成长长整型，可以用来防止数字过大存不下qaq。
    n--;
    m--;
    cout<<(m-1)/7-(n-1)/7<<endl;//[m,n)区间内有多少个整数i满足i%x=y，公式如下：
    m--,n--,(m-y)/x-(n-y)/x
    return 0;
}

B-Relic Discobery

链接：https://ac.nowcoder.com/acm/contest/5902/B
来源：牛客网

Recently, paleoanthropologists have found historical remains on an island in the Atlantic Ocean. The most inspiring thing is that they excavated in a magniﬁcent cave and found that it was a huge tomb. Inside the construction, researchers identiﬁed a large number of skeletons, and funeral objects including stone axe, livestock bones and murals. Now, all items have been sorted, and they can be divided into N types. After they were checked attentively, you are told that there are Ai items of the i-th type. Further more, each item of the i-th type requires Bi million dollars for transportation, analysis, and preservation averagely. As your job, you need to calculate the total expenditure.

输入描述:

The ﬁrst line of input contains an integer T which is the number of test cases. For each test case, the ﬁrst line contains an integer N which is the number of types. In the next N lines, the i-th line contains two numbers Ai and Bi as described above. All numbers are positive integers and less than 101.

输出描述:

For each case, output one integer, the total expenditure in million dollars.

示例1

输入

复制

输出

复制

#include<bits/stdc++.h>

using namespace std;
int t,n;
int a,b;

int main()
{
    cin>>t;
    while(t--)
    {
        int ans=0;
        cin>>n;
        while(n--)//相乘之和即可
        {
            cin>>a>>b;
            ans+=a*b;
        }
        cout<<ans<<endl;
    }
    return 0;
}

洛谷篇（算法1-1）模拟与高精度

P4924[1007]魔法少女小Scarlet

#include<bits/stdc++.h>

using namespace std;//简单模拟即可
int g[510][510],tot,f[510][510];//f数组充当临时数组 

int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            g[i][j]=++tot;
    for(int i=1;i<=m;i++) {
        int a,b,r,opt;
        scanf("%d %d %d %d",&a,&b,&r,&opt);//下面的分析都是针对那一块矩阵而言的 
        if(opt==0) {//第i行第j个 变成倒数第i列第j个 顺时针 
            for(int i=a-r;i<=a+r;i++)
                for(int j=b-r;j<=b+r;j++)
                    f[a-b+j][a+b-i] = g[i][j];
            for(int i=a-r;i<=a+r;i++)
                for(int j=b-r;j<=b+r;j++)
                    g[i][j] = f[i][j];
        }
        else { //第i行第j个 变成第i列倒数第j个 逆时针 
             for(int i=a-r;i<=a+r;i++)
                for(int j=b-r;j<=b+r;j++)
                    f[a+b-j][b-a+i] = g[i][j];
             for(int i=a-r;i<=a+r;i++)
                for(int j=b-r;j<=b+r;j++)
                    g[i][j] = f[i][j];
        }
    }
    for(int i=1;i<=n;i++) {//输出结果咯 
        for(int j=1;j<=n;j++)
            printf("%d ",g[i][j]);
        printf("\n");
    }
    return 0;
}

P1328生活大爆炸石头剪刀布

#include<bits/stdc++.h>

using namespace std;
int n,na,nb;
int a[220],b[220];
int numa,numb;

int main()
{
    cin>>n>>na>>nb;//理清思路，简单模拟即可
    for(int i=0;i<na;i++) cin>>a[i];
    for(int i=0;i<nb;i++) cin>>b[i];
    for(int i=0;i<n;i++)
    {
        if(a[i%na]==0&&(b[i%nb]==2||b[i%nb]==3)) numa++;
        else if(a[i%na]==1&&(b[i%nb]==0||b[i%nb]==3)) numa++;
        else if(a[i%na]==2&&(b[i%nb]==1||b[i%nb]==4)) numa++;
        else if(a[i%na]==3&&(b[i%nb]==2||b[i%nb]==4)) numa++;
        else if(a[i%na]==4&&(b[i%nb]==0||b[i%nb]==1)) numa++;
        else if(a[i%na]==b[i%nb]) numa=numa,numb=numb;
        else numb++;
    }
    cout<<numa<<" "<<numb;
    return 0;
}