1140. Look-and-say Sequence (20)Look-and-say sequence is a sequence of integers as the following: D, D1, D111, D113, D11231, D112213111, ...
where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is one D in the 1st number, and hence it is D1; the 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to 11), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, two 1's, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digit D. Input Specification: Each input file contains one test case, which gives D (in [0, 9]) and a positive integer N (<=40), separated by a space. Output Specification: Print in a line the Nth number in a look-and-say sequence of D. Sample Input:1 8
Sample Output:1123123111 2018.3.18 PAT 94~ 这题开始读题目读了10分钟。。。没看懂，先做了后面3题，回来在做了这题。在这道题目上差不多花了1小时。。。 输入样例，除了第一个，其他2个2个看。第一个只有D，所以第二个D1，这里D表示字符，1表示D的个数。第三个D1 11 ,D表示字符，紧跟的1表示D的个数，然后下一个1表示字符1，紧跟后面的1表示字符1的个数...
为什么考试最后一点超时，现在提交就过了呢。。。虽然是345ms。。。 #include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
char change(int x){//数字字符与字符阿斯克吗相差 48
return x+48;
}
int main(){
int n,i,j,cou;
string ans;
cin>>ans>>n;
if(n==1){
cout<<ans;return 0;
}
n--;
while(n--){
string tmp="";
for(i=0;i<ans.size();i++){
tmp=tmp+ans[i];
cou=1;
char c=ans[i];
for(j=i+1;j<ans.size();j++){
if(ans[j]!=c){
break;
}
else{
cou++;//计算连续相同的有几个
}
}
i=i+cou-1;//跳过连续相同的
char aa=change(cou) ;//把个数转换为字符
tmp=tmp+aa;
}
ans=tmp;
}
cout<<ans;
}
靴习一个 to_string 记得在DEV-工具-编译选项-编译器-[勾选 编译时加入以下命令] 填入 -std=c++11 （小写c） #include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
//char change(int x){//数字字符与字符阿斯克吗相差 48
// return x+48;
//}
int main(){
int n,i,j,cou;
string ans;
cin>>ans>>n;
if(n==1){
cout<<ans;return 0;
}
n--;
while(n--){
string tmp="";
for(i=0;i<ans.size();i++){
tmp=tmp+ans[i];
cou=1;
char c=ans[i];
for(j=i+1;j<ans.size();j++){
if(ans[j]!=c){
break;
}
else{
cou++;//计算连续相同的有几个
}
}
i=i+cou-1;//跳过连续相同的
// char aa=change(cou) ;//把个数转换为字符
tmp=tmp+to_string(cou);//aa;
}
ans=tmp;
}
cout<<ans;
}
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