A1071

#include<bits/stdc++.h>
using namespace std;
int main(){
    int i=1;
    int fi=1;
    int fnext=1;

    int k;
    cin>>k;

    while (i<k){
         //fi,fnext --------> fnext,fi+fnext
        int fnextnext = fi + fnext;
        fi =fnext ;
           fnext=fnextnext;
        i++;
    }
    cout<<fi;
    return 0;
    }

A1077

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin>>n;
    int i=0;
    int answer=0;
    while (i<n){
        //i从0~~~n-1 重复执行n次
        int x;
        cin>>x; //读入四位数x
        int ge =x/1   %10;
        int shi=x/10  %10;
        int bai=x/100 %10;
        int qian=x/1000%10;

        if (ge>shi+bai+qian) 
            answer = answer + 1;
        i++;
        //最后一次执行之后 i变成n
    }
    cout<<answer;
    return 0;
}

A1078

#include<bits/stdc++.h>
using namespace std;
int main() {
int n;
cin>>n;
int i=1;
int fi0=1, fi1=1, fi2=2;
double ans = 0;
while (i<=n) {
ans = ans +1.0*fi2/fi1;
fi0 = fi1;
fi1 = fi2;
fi2 = fi1+fi0;
i = i + 1;
}
cout<<fixed<<setprecision(4)<<ans<<endl;
return 0;
}

A1079

#include<stdio.h>
int main(){
    int n; double ans = 0.0;
    scanf ("%d", &n);
    for (int i = 1;i <= n;++i)
        if (i % 2 == 0) ans -= 1.0/i;
        else ans += 1.0/i;
    printf ("%.4lf", ans);
}

A1080

#include<iostream>
using namespace std;
int main(){
int a,b,c;
cin>>a>>b>>c;
for(int x=2;x<=a;x++)
if(a%x==b%x && a%x==c%x)
{
cout<<x<<endl;
return 0;
}

A1083

#include<iostream>
using namespace std;
int main(){
    int a,b;
    cin>>a>>b;
    int s=1;
    for(int i=1;i<=b;i++){
        s=s*a%7;
    }
    if(s==0) cout<<"Sunday";
    if(s==1) cout<<"Monday";
    if(s==2) cout<<"Tuesday";
    if(s==3) cout<<"Wednesday";
    if(s==4) cout<<"Thursday";
    if(s==5) cout<<"Friday";
    if(s==6) cout<<"Saturday";
    return 0;
}

A1084

#include<iostream>
using namespace std;
int main(){
    int a,b;
    cin>>a>>b;
    int s=1;
    for(int i=1;i<=b;i++){
        s=s*a%1000;
    }
    if(s<100)cout<<0;
    if(s<10) cout<<0;
    cout<<s;
    return 0;
}

A1085

#include<iostream>
using namespace std;
int main(){
    double h;
    cin>>h;
    cout<<3*h-h/256<<endl;
    cout<<h/1024<<endl;
    return 0;
}

A1110

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int a[100000],n,x,ans=0;
    cin>>n;

    for (int i=1;i<=n;i++){
        cin>>a[i];
    }
    cin>>x;
    for(int i=1;i<=n;i++){
        cin>>a[i];

        if(a[i]==x){
            ans=i;
            break;
        }
    }

    cout<<ans;
    return 0;
}

A1107

#include<stdio.h>  
  
int main()  
{  
    int L, M, i, j, n;  
    int a[10001], b[10001];  
    scanf("%d %d",&L, &M);   //输入L和M  
    n = M*2;      //循环输入b数组0~n的数据  
    for(i=0; i<n; i+=2)  
    {  
        scanf("%d %d", &b[i], &b[i+1]);  
    }  
    for(i=0; i<=L; i++)     //循环给a数组L个元素赋值  
    {  
        a[i] = i;  
    }  
    int r, s;  
    for(i=0; i<n; i+=2)  //遍历访问数组b的各个区间  
    {  
        r = b[i];   //区间起始点  
        s = b[i+1]; //区间终点  
        for(j=r; j<=s; j++) //把数组b各个区间内元素在数组a中映射为0  
        {  
            a[j] = -1;  
        }  
    }  
    int k=0;    //k用来统计不是0的个数，k赋值为1（数组a中原本包含一个0，此时加上）  
    for(i=0; i<=L; i++)  
    {  
        if(a[i] != -1)  
        {  
            k++;    //用来统计非0的个数，即不在数组b区间内的数的个数  
        }  
    }  
    printf("%d", k);  
  
    return 0;  
}

A1118

#include<bits/stdc++.h>
using namespace std;

const int N=100005;
int x[N],y[N],xlen[N],ylen[N];

int main()
{
int n; cin>>n;
for(int i=1;i<=n;i++){
cin>>x[i]>>y[i]>>xlen[i]>>ylen[i];
}
int px,py; cin>>px>>py;
int ans=-1;
for(int i=1;i<=n;i++){
if(px>=x[i] && px<=x[i]+xlen[i] && py>=y[i] && py<=y[i]+ylen[i])
ans = i;
}
cout<<ans;
return 0;
}

A1124

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[10001][10001];
int b[10001][10001];
int ans[10001][10001];
int now;
int tot;
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>a[i][j];
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>b[i][j];
            ans[i][j]=a[i][j]+b[i][j];
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cout<<ans[i][j]<<" ";
        }
        cout<<endl;
    }
    return 0;
}