第十一章模拟 4 AcWing 1515. U 形 Hello World-蒲公英云

第十一章模拟 4 AcWing 1515. U 形 Hello World

原题链接

AcWing 1515. U 形 Hello World

算法标签

模拟

思路

计算符合题意得n1, n2, n3, 模拟填入矩阵输出

如何计算符合题意得n1, n2, n3

在这里插入图片描述

代码

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define int long long
#define xx first
#define yy second
#define ump unordered_map
#define us unordered_set
#define pq priority_queue
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>=b;--i)
using namespace std;
typedef pair<int, int> PII;
const int N=85, inf=0x3f3f3f3f3f3f3f3f, mod=1e9+7;
const double Exp=1e-8;
//int t, n, m, cnt, ans;
char g[N][N], c[N];
string s;
int n, n1, n2, k;
inline int rd(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
void put(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar(x%10^48);
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    scanf("%s", c);
    s=c;
    n=s.size();
    int n1=(n+2)/3;
    int n2=n+2-n1*2;
    k=0;
    rep(i, 0, n1){
        g[i][0]=s[k++];
    }
    rep(i, 1, n2){
        g[n1-1][i]=s[k++];
    }
    Rep(i, n1-2, 0){
        g[i][n2-1]=s[k++];
    }
    rep(i, 0, n1){
        rep(j, 0, n2){
            if(g[i][j]){
                printf("%c", g[i][j]);
            }else{
                printf(" ");
            }
        }
        puts(" ");
    }
    return 0;
}