新手天翼
计算机内信息编码采用二进制。现给定一串由小写字母构成的单词，请根据后面的编码原则，输出对应的二进制编码。已知编码长度n和编码的前n-5位，后5位的编码原则：a是00001，b是00010，c是00011，d是00100，以此类推，即后一个的编码是在前一个编码的基础上加1。
输入共3行，第一行一个正整数n，表示编码长度；第二行给定前n-5位的二进制编码；第三行是m个小写字母构成的单词。
m行，每行对应一个小写字母的二进制编码。
8 110 will
11010111 11001001 11001100 11001100
8<=n<=16，1<=m<=30
本人90分代码如下，找错：
#include<iostream>
#include<string>
using namespace std;
string a1[100]={"00001","00010","00011","00100","00101","00110","00111","01000","01001","01010","01011","01100","01101","01110","01111","10000"
,"10001","10010","10011","10100","10101","10110","10111","11000","11010","11011"};
char b1[100]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
string b;
int a,n;
int main()
{
cin>>n;
cin>>a>>b;
for(int i=0;i<b.size();i++)
{
for(int j=0;j<26;j++)
{
if(b[i]==b1[j])
{
cout<<a<<a1[j]<<endl;
break;
}
}
}
return 0;
}
求大佬解答
新手天翼
ai课堂有
缔造者之神
1
修练者
AI课堂吗？
时梓繁在2018-08-28 09:22:42追加了内容
是的话：
#include<bits/stdc++.h>
using namespace std;
int d[201][201],dis[201],f[201];
int main()
{
int n,a,b;
cin>>n>>a>>b;
for (int i=1;i<=n;++i)
{
for (int j=1;j<=n;++j)
d[i][j]=1000000;
d[i][i]=0;
}
int k;
for (int i=1;i<=n;++i)
{
scanf("%d",&k);
if (i+k<=n) d[i][i+k]=1;
if (i-k>=1) d[i][i-k]=1;
}
for (int i=1;i<=n;++i)
dis[i]=d[a][i];
int minn,t;
for (int i=1;i<=n;++i)
{
minn=1000000;
for (int j=1;j<=n;++j)
if (f[j]==0&&dis[j]<minn)
{
minn=dis[j];
t=j;
}
f[t]=1;
for (int j=1;j<=n;++j)
dis[j]=min(dis[j],dis[t]+d[t][j]);
}
if (dis[b]!=1000000)
cout<<dis[b];
else
cout<<-1;
return 0;
}
时梓繁在2018-08-28 09:24:26追加了内容
不是的话：请问题号是什么
缔造者
1
中级光能
我将你不对的地方修改了一下
（也就是你代码有误的地方）
但改完后依然是90分
此为代码：
#include<iostream>
#include<string>
using namespace std;
string a1[100]={"00001","00010","00011","00100","00101","00110","00111","01000","01001","01010","01011","01100","01101","01110","01111","10000","10001","10010","10011","10100","10101","10110","10111","11000","11001","11010"};
char b1[100]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
string b;
int a,n;
int main()
{
cin>>n;
if(n>5)
cin>>a;
cin>>b;
for(int i=0;i<b.size();i++)
{
for(int j=0;j<26;j++)
{
if(b[i]==b1[j])
{
cout<<a<<a1[j]<<endl;
break;
}
}
}
return 0;
}
请自己再想想！！！
望采纳！！！
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