AtCoder Beginner Contest 072

T1:
A-B B>A時輸出0

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long ll;
int main()
{
    ll x,t;
    scanf("%lld%lld",&x,&t);
    if(t>=x) cout<<"0"<<endl;
    else cout<<x-t<<endl;
    return 0;
}

T2:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long ll;
const int N=1e5+10;
char ch[N];
int main()
{
    scanf("%s",ch+1);
    int n=strlen(ch+1);
    for(int i=1;i<=n;i+=2) {
        cout<<ch[i];
    }
    cout<<endl;
    return 0;
}

T3:
記錄每種可能,取最大值即可

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long ll;
const int N=1e5+10;
int ans[N],n,tot=0;
int main()
{
    memset(ans,0,sizeof(ans));
    scanf("%d",&n);
    for(int x,i=1;i<=n;i++) {
        scanf("%d",&x);
        if(x!=0) {ans[x+1]++;ans[x]++;ans[x-1]++;}
        else {ans[x]++;ans[x+1]++;tot++;}
    }
    int cnt=0;
    fo(i,0,N-1) cnt=max(cnt,ans[i]);
    cout<<cnt<<endl;
    return 0;
}

T4:模擬即可:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=1e5+10;
int a[N],n,ans=0;
int main()
{
    scanf("%d",&n);
    fo(i,1,n) scanf("%d",&a[i]);
    for(int i=1;i<=n;i++) {
        if(a[i]==i) {
            if(i==n) swap(a[i],a[i-1]);
            else swap(a[i],a[i+1]);
            ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}