Martian Addition（模拟）
Time Limit: 2 Seconds Memory Limit: 65536 KB
In the 22nd Century, scientists have discovered intelligent residents live on the Mars. Martians are very fond of mathematics. Every year, they would hold an Arithmetic Contest on Mars (ACM). The task of the contest is to calculate the sum of two 100-digit numbers, and the winner is the one who uses least time. This year they also invite people on Earth to join the contest.
As the only delegate of Earth, you're sent to Mars to demonstrate the power of mankind. Fortunately you have taken your laptop computer with you which can help you do the job quickly. Now the remaining problem is only to write a short program to calculate the sum of 2 given numbers. However, before you begin to program, you remember that the Martians use a 20-based number system as they usually have 20 fingers.
Input:
You're given several pairs of Martian numbers, each number on a line.
Martian number consists of digits from 0 to 9, and lower case letters from a to j (lower case letters starting from a to present 10, 11, ..., 19).
The length of the given number is never greater than 100.
Output:
For each pair of numbers, write the sum of the 2 numbers in a single line.
Sample Input:Sample Output:1234567890 abcdefghij 99999jjjjj 9999900001bdfi02467j iiiij00000
【分析】就是二十进制运算....我也不知道为什么 写了一个下午...不知道最近什么情况，写什么都要写很久很久....
注意特例，01+0000000000001；0+0；就是前导很多0的时候要处理一下。然后就是，长度不一样的话，前补0；
其他好像没什么要注意的。反正就是菜得不行，找错误找了那么久。。好烦。。。
【代码】
#include<bits/stdc++.h>
using namespace std;const int maxn=110;
int a[maxn],b[maxn];string add(string x,string y)
{memset(a,0,sizeof(a));memset(b,0,sizeof(b));int len1=x.length(),len2=y.length();for(int i=0;i<len1;++i)a[len1-1-i]=(isdigit(x[i]))?x[i]-'0':x[i]-'a'+10;for(int i=0;i<len2;++i)b[len2-1-i]=(isdigit(y[i]))?y[i]-'0':y[i]-'a'+10;int len=max(len1,len2);string ans="";for(int i=0;i<len;++i){a[i]+=b[i];a[i+1]+=a[i]/20;a[i]%=20;}if(a[len])len++;for(int i=len-1;i>=0;--i){if(a[i]<10)ans+=a[i]+'0';else ans+=a[i]-10+'a';}return ans;
}int main()
{string s1,s2;while(cin>>s1>>s2){printf("%s\n",add(s1,s2).c_str());}return 0;
}
Parencodings（模拟）
Time Limit: 2 Seconds Memory Limit: 65536 KB
Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:
- By an integer sequence P = p1 p2 ... pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
- By an integer sequence W = w1 w2 ... wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
S (((()()())))
P-sequence 4 5 6666
W-sequence 1 1 1456
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.
Output
The output consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.
Sample Input
2
6
4 5 6 6 6 6
9
4 6 6 6 6 8 9 9 9
Sample Output
1 1 1 4 5 6
1 1 2 4 5 1 1 3 9
【分析】P序列：第i个右括号前左括号的数量 ；W序列：第i个右括号前成功匹配的左括号的数量；题中红字部分说明括号是成对匹配好的，不存在前面有很多个左括号然后最后只有一个右括号。所以，可以先把这个序列还原，再做。有点麻烦。
【代码】
#include<bits/stdc++.h>
using namespace std;
char s[100];
int a[100];
int main()
{int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);a[0]=0;for(int i=1;i<=n;i++)scanf("%d",&a[i]);int len=0;for(int i=1;i<=n;i++){for(int j=a[i-1];j<a[i];j++)s[len++]='(';if(a[i]-a[i-1]>=0)s[len++]=')';}//for(int i=0;i<len;i++)cout<<s[i];int ans=0,num,flag=0;for(int i=0;i<len;i++){if(s[i]==')'){num=0;for(int j=i;j>=0;j--){if(s[j]=='(')ans--;else if(s[j]==')')ans++,num++;if(ans==0)break;}if(flag)printf(" ");flag=1;printf("%d",num);}}puts("");}return 0;}