「CF435C」Cardiogram
In this problem, your task is to use ASCII graphics to paint a cardiogram.
A cardiogram is a polyline with the following corners:
Your task is to paint a cardiogram by given sequence ai.
The first line contains integer n (2 ≤ n ≤ 1000). The next line contains the sequence of integersa1, a2, …, an (1 ≤ ai ≤ 1000). It is guaranteed that the sum of all ai doesn’t exceed 1000.
Print max |yi - yj| lines (where yk is the y coordinate of the k-th point of the polyline), in each line print 
characters. Each character must equal either « / » (slash), « \ » (backslash), « » (space). The printed image must be the image of the given polyline. Please study the test samples for better understanding of how to print a cardiogram.
Note that in this problem the checker checks your answer taking spaces into consideration. Do not print any extra characters. Remember that the wrong answer to the first pretest doesn’t give you a penalty.
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5 3 1 2 5 1 |
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1 2 |
3 1 5 1 |
Due to the technical reasons the answers for the samples cannot be copied from the statement. We’ve attached two text documents with the answers below.
题解
比较麻烦的模拟。。。
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#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #define ll long long #define inf 0x7fffffff using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x*=10;x+=ch-'0';ch=getchar();} return x*f; } int n,tot,mx=1000,mn=1000,x,y=1000; int a[1005]; int b[2005][2005]; int main() { n=read(); for(int i=1;i<=n;i++) a[i]=read(),tot+=a[i]; for(int i=1;i<=n;i++) { x++; if(i&1) { a[i]--; b[x][y]=1; while(a[i]--) { x++;y++; b[x][y]=1; } mx=max(mx,y); } else { a[i]--; b[x][y]=2; while(a[i]--) { x++;y--; b[x][y]=2; } mn=min(mn,y); } } for(int i=mx;i>=mn;i--) { for(int j=1;j<=tot;j++) if(b[j][i]==1)printf("/"); else if(b[j][i]==2)printf("\\"); else printf(" "); printf("\n"); } return 0; } |
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