标签：数学 模拟 codeforces
In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
The first line of the input contains integer number n (1?≤?n?≤?105), the number of coordinates in the test. Then there follow n lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 .
Write n lines, each line should contain a cell coordinates in the other numeration system.
2
R23C55
BC23
BC23 R23C55 题目意思： 简单来说就是10进制和26进制的相互转换，给出RXCY转换长YX，其中Y为26进制，模拟题目，不难但是各种陷阱 各种多，看自己的人品了，我都不知道wa了多少次。。。。，还得注意判断是那种转换，是RXCY-->XY?还是 XY—>RXCY AC代码：/** *@xiaoran *题意意思，26进制的转换 */ #include<iostream> #include<cstdio> #include<map> #include<cstring> #include<string> #include<algorithm> #include<queue> #include<vector> #include<stack> #include<cstdlib> #include<cctype> #include<cmath> #define LL long long using namespace std; void IntToSta(char *s){ int sr,sc,len; sr=sc=0; len=strlen(s); for(int i=0;i<len;i++){ if(isalpha(s[i])){//是字母 sc=sc*26+(s[i]-'A'+1); } if(isdigit(s[i])){//是数字 sr=sr*10+(s[i]-'0'); } } printf("R%dC%d\n",sr,sc); } void StaToInt(char *s){ int sr,sc,len; //sr=sc=0; len=strlen(s); sscanf(s,"R%dC%d",&sr,&sc); //printf("%d %d\n",sr,sc); int res=sc,mes,k=0; //res=sc/26; mes=sc%26; char ss[58]; while(res){ if(res%26==0){ ss[k++]='A'+res%26-1+26; res/=26; res-=1; } else{ ss[k++]='A'+res%26-1; //printf("%c",'A'+res%26); res/=26; } } //ss[k]='A'+res-1; for(int i=k-1;i>=0;i--) printf("%c",ss[i]); printf("%d\n",sr); //printf("R%dC%d\n",sr,sc); } int main() { int n; char s[100]; scanf("%d",&n); while(n--){ scanf("%s",s); int ok1=0,ok2=0; ok1=s[0]=='R'&&isdigit(s[1]); for(int i=0;i<strlen(s)-1;i++){ if(s[i]=='C'&&isdigit(s[i+1])) ok2=1; } //ok1&ok2判断选择哪个函数 if(ok1&&ok2) StaToInt(s); else IntToSta(s); } return 0; }
标签：数学 模拟 codeforces
原文地址：http://blog.csdn.net/fool_ran/article/details/44197063
