杭电oj~1228 map=_=
A + B
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15387 Accepted Submission(s): 9162
Problem Description
读入两个小于100的正整数A和B,计算A+B.
需要注意的是:A和B的每一位数字由对应的英文单词给出.
需要注意的是:A和B的每一位数字由对应的英文单词给出.
Input
测试输入包含若干测试用例,每个测试用例占一行,格式为"A + B =",相邻两字符串有一个空格间隔.当A和B同时为0时输入结束,相应的结果不要输出.
Output
对每个测试用例输出1行,即A+B的值.
Sample Input
one + two =
three four + five six =
zero seven + eight nine =
zero + zero =
Sample Output
3
90
96
Source
#include<iostream> #include<map> #include<cstring> using namespace std; map<string,int>m; int main() { m["zero"]=0; m["one"]=1; m["two"]=2; m["three"]=3; m["four"]=4; m["five"]=5; m["six"]=6; m["seven"]=7; m["eight"]=8; m["nine"]=9; m["+"]=10; m["="]=11; int cnt[2]={0}; char a[20]; int k=0;int z=0; int sum[2]={0}; while(cin>>a) { int d=m[a]; if(d<10) cnt[k++]=d; if(d==10) { if(k==1) sum[0]=cnt[0]; else if(k==2) sum[0]=10*cnt[0]+cnt[1]; k=0; } if(d==11) { if(k==1) sum[1]=cnt[0]; else if(k==2) sum[1]=10*cnt[0]+cnt[1]; k=0; if(sum[0]+sum[1]==0) break; else cout<<sum[0]+sum[1]<<endl; } } return 0; }