ZOJ 2971 Give Me the Number
2017-02-17 21:28
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Numbers in English are written down in the following way (only numbers less than 109are considered). Number abc,def,ghi is written as "[abc] million [def] thousand [ghi]". Here "[xyz] "
means the written down number xyz .
In the written down number the part "[abc] million" is omitted if abc = 0 , "[def] thousand" is omitted if def = 0 , and "[ghi] " is omitted if ghi = 0 . If the whole number is equal
to 0 it is written down as "zero". Note that words "million" and "thousand" are singular even if the number of millions or thousands respectively is greater than one.
Numbers under one thousand are written down in the following way. The number xyz is written as "[x] hundred and [yz] . ( If yz = 0 it should be only [x] hundred. Otherwise if y = 0 it
should be only [x] hundred and [z].) Here "[x] hundred and" is omitted if x = 0 . Note that "hundred" is also always singular.
Numbers under 20 are written down as "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", and "nineteen" respectively. Numbers
from 20 to 99 are written down in the following way. Number xy is written as "[x0][y] ", and numbers divisible by ten are written as "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", and "ninety"
respectively.
For example, number 987,654,312 is written down as "nine hundred and eighty seven million six hundred and fifty four thousand three hundred and twelve", number 100,000,037 as "one hundred million thirty seven", number 1,000 as
"one thousand". Note that "one" is never omitted for millions, thousands and hundreds.
Give you the written down words of a number, please give out the original number.
Input
Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 1900) which is the number of test cases. It will be followed by T consecutive test cases.
Each test case contains only one line consisting of a sequence of English words representing a number.
Output
For each line of the English words output the corresponding integer in a single line. You can assume that the integer is smaller than 109.
Sample Input
Sample Output
简单模拟,按照数字的规则写就行#include<map>
#include<ctime>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e3 + 10;
int T;
char s
;
string g[20] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
string f[8] = { "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };
map<string, int> M;
int main()
{
rep(i, 0, 19) M[g[i]] = i;
rep(i, 0, 7) M[f[i]] = i * 10 + 20;
for (inone(T),getchar(); T--;)
{
gets(s);
int ans = 0, now = 0;
string x = "";
for (int i = 0; s[i]; i++)
{
if (s[i] == ' ')
{
if (x == "hundred") now = now * 100;
if (x == "million") { ans += now * 1000000; now = 0; }
if (x == "thousand") { ans += now * 1000; now = 0; }
now += M[x]; x = "";
}
else x = x + s[i];
}
if (x == "hundred") now = now * 100;
if (x == "million") { ans += now * 1000000; now = 0; }
if (x == "thousand") { ans += now * 1000; now = 0; }
now += M[x]; x = "";
printf("%d\n", ans + now);
}
return 0;
}
means the written down number xyz .
In the written down number the part "[abc] million" is omitted if abc = 0 , "[def] thousand" is omitted if def = 0 , and "[ghi] " is omitted if ghi = 0 . If the whole number is equal
to 0 it is written down as "zero". Note that words "million" and "thousand" are singular even if the number of millions or thousands respectively is greater than one.
Numbers under one thousand are written down in the following way. The number xyz is written as "[x] hundred and [yz] . ( If yz = 0 it should be only [x] hundred. Otherwise if y = 0 it
should be only [x] hundred and [z].) Here "[x] hundred and" is omitted if x = 0 . Note that "hundred" is also always singular.
Numbers under 20 are written down as "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", and "nineteen" respectively. Numbers
from 20 to 99 are written down in the following way. Number xy is written as "[x0][y] ", and numbers divisible by ten are written as "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", and "ninety"
respectively.
For example, number 987,654,312 is written down as "nine hundred and eighty seven million six hundred and fifty four thousand three hundred and twelve", number 100,000,037 as "one hundred million thirty seven", number 1,000 as
"one thousand". Note that "one" is never omitted for millions, thousands and hundreds.
Give you the written down words of a number, please give out the original number.
Input
Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 1900) which is the number of test cases. It will be followed by T consecutive test cases.
Each test case contains only one line consisting of a sequence of English words representing a number.
Output
For each line of the English words output the corresponding integer in a single line. You can assume that the integer is smaller than 109.
Sample Input
3 one eleven one hundred and two
Sample Output
1 11102
简单模拟,按照数字的规则写就行#include<map>
#include<ctime>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e3 + 10;
int T;
char s
;
string g[20] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
string f[8] = { "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };
map<string, int> M;
int main()
{
rep(i, 0, 19) M[g[i]] = i;
rep(i, 0, 7) M[f[i]] = i * 10 + 20;
for (inone(T),getchar(); T--;)
{
gets(s);
int ans = 0, now = 0;
string x = "";
for (int i = 0; s[i]; i++)
{
if (s[i] == ' ')
{
if (x == "hundred") now = now * 100;
if (x == "million") { ans += now * 1000000; now = 0; }
if (x == "thousand") { ans += now * 1000; now = 0; }
now += M[x]; x = "";
}
else x = x + s[i];
}
if (x == "hundred") now = now * 100;
if (x == "million") { ans += now * 1000000; now = 0; }
if (x == "thousand") { ans += now * 1000; now = 0; }
now += M[x]; x = "";
printf("%d\n", ans + now);
}
return 0;
}
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