3.17模拟总结

今天终于弄明白了lemon评测的文件格式！

T1

 

经典的一道题，贪心：对于第一个元素，只能由第二个元素给他，给完之后，第二个元素就变成了第一个元素，一直循环下去就行

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
using namespace std;
int n,a[105]; 
int sum,ans;
int main(){
    freopen("Playcard.in","r",stdin);
    freopen("Playcard.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        scanf("%d",a+i);
        sum += a[i];
    }
    sum /= n;
    for(int i=1;i<=n;++i){
        int x;
        if(a[i] != sum){
            ans++;
            x = a[i] - sum;
            a[i] = sum;
            a[i+1] += x; 
        }
    }
    printf("%d",ans);
}

T2

 

 贪心，每次找到第一个单调减的数字，删掉

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
using namespace std;
int main(){
    freopen("delete.in","r",stdin);
    freopen("delete.out","w",stdout);
    string a;
    int s;
    cin >> a >> s;
    int len = a.length();
    for(int i=1;i<=s;++i){
        for(int j=0;j<len;++j){
            if(a[j] > a[j+1]){
                for(int k=j;k<len-1;++k)
                    a[k] = a[k+1];
                break;
            }
        }
    }
    int flag = 1;
    for(int i=0;i<len-s;++i){
        if(a[i] != '0')flag = 0;
        if(!flag)putchar(a[i]);
    }
    if(flag)putchar('0');
}

T3

 

 要求最长不上升子序列个数，就是求最长上升序列的长度，树状数组维护dp即可

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
using namespace std;
const int N = 30000 + 5;
int c[N],a[1005],n,ans,mx;
inline int lowbit(int x){
    return x & (-x);
}
inline void update(int i,int x){
    for(;i<=mx;i+=lowbit(i))c[i] = max(c[i],x);
}
inline int query(int i){
    int res = 0;
    for(;i;i-=lowbit(i))res = max(res,c[i]);
    return res;
}
int main(){
    freopen("missile.in","r",stdin);
    freopen("missile.out","w",stdout);
    int x;
    while(scanf("%d",&x) == 1){
        a[++n] = x;
        mx = max(mx,x);
    }
    for(int i=1;i<=n;++i){
        int now = query(a[i])+1;
        ans = max(ans,now);
        update(a[i]+1,now);
    }
    printf("%d",ans);
}

T4

 

 这个就是线段覆盖问题，贪心的板子，注意这里的区间左闭右开，我就当成闭区间错了（

按右端点排序，如果当前左端点不小于之前最右端点，就能加入

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
using namespace std;
struct qwq{
    int st,ed;
    bool operator <(qwq x)const{
        if(ed != x.ed)return ed < x.ed;
        return st > x.st;
    } 
}a[1005];
int main(){
    freopen("act.in","r",stdin);
    freopen("act.out","w",stdout);
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%d%d",&a[i].st,&a[i].ed);
    sort(a+1,a+1+n);
    int now_end = -1,ans = 0;
    for(int i=1;i<=n;++i){
        if(a[i].st >= now_end){
            now_end = a[i].ed;
            ans++;
        }
    }
    printf("%d",ans);
}

T5

 

 大体上和上一道题一样，右端点排序，但这回是闭区间，而且条件变成当前区间左端点不超过之前最右端点

#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<vector>
#include<string>
using namespace std;
struct qwq{
    int x,y;
    bool operator <(qwq b)const{
        return y < b.y;
    }
}a[10005];
int main(){
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout); 
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%d%d",&a[i].x,&a[i].y);
    sort(a+1,a+1+n);
    int now = a[1].y;
    int ans = 1;
    for(int i=2;i<=n;++i){
        if(now >= a[i].x)continue;
        ans++;
        now = a[i].y;
    }
    printf("%d",ans);
}