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Total Submission(s): 1391 Accepted Submission(s): 577
Problem Description
Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single line with several words. There will be at most 1000 characters in a line.
Each test case contains a single line with several words. There will be at most 1000 characters in a line.
Output
For each test case, you should output the text which is processed.
Sample Input
3 olleh !dlrow m'I morf .udh I ekil .mca
Sample Output
hello world! I'm from hdu. I like acm.HintRemember to use getchar() to read '\n' after the interger T, then you may use gets() to read a line and process it.
Author
Ignatius.L
#include <string.h>
int main()
{
int i=0,n,length,t,j;
char ch,p[1001],temp;
scanf("%d",&n);
ch=getchar();
while(n--)
{
gets(p);
length=strlen(p);
t=0;
for(i=0;i<=length;i++)
{
if(p[i]==' '||p[i]=='\0')
{
for(j=t;j<(i+t)/2;j++)
{
temp=p[j];
p[j]=p[i-j+t-1];
p[i-j+t-1]=temp;
}
t=i+1;
}
}
printf("%s\n",p);
}
return 0;
}
]]>
#include <string.h>
#include <ctype.h>
char * index( const char *s, int c);
/*
** State machine for romantoint
**
** M D C L X V I
** 0: +1000/0 +500/0 +100/1 +50/0 +10/2 +5/0 +1/3
** 1: +800/0 -- +100/0 +50/0 +10/2 +5/0 +1/3
** 2: -- -- +80/0 +30/0 +10/2 +5/0 +1/3
** 3: -- -- -- -- +8/0 +3/0 +1/3
**
*/
int FromRoman(char *s)
{
int n=0;
int prev=0;
while (*s!=' '&&*s!=NULL) {
switch (toupper(*s))
{
case 'M': n += 1000 - prev*2;
prev=0;
break;
case 'D': n += 500 - prev*2;
prev=0;
break;
case 'C': n += 100;
if (prev < 100) n -= 2*prev;
prev=100;
break;
case 'L': n += 50;
if (prev < 50) n -= 2*prev;
prev=0;
break;
case 'X': n += 10;
if (prev < 10) n -= 2*prev;
prev=10;
break;
case 'V': n += 5;
if (prev < 5) n -= 2*prev;
prev=0;
break;
case 'I': n += 1;
prev=1;
break;
}
s++;
}
return n;
}
void PrintRoman(int n)
{
if (n<=0 || n>4999)
{
printf("out of range exception\n");
return;
}
while (n/1000)
{
printf("M");
n -= 1000;
}
switch(n/100)
{
case 0: break;
case 1: printf("C"); break;
case 2: printf("CC"); break;
case 3: printf("CCC"); break;
case 4: printf("CD"); break;
case 5: printf("D"); break;
case 6: printf("DC"); break;
case 7: printf("DCC"); break;
case 8: printf("DCCC"); break;
case 9: printf("CM"); break;
}
n %= 100;
switch(n/10)
{
case 0: break;
case 1: printf("X"); break;
case 2: printf("XX"); break;
case 3: printf("XXX"); break;
case 4: printf("XL"); break;
case 5: printf("L"); break;
case 6: printf("LX"); break;
case 7: printf("LXX"); break;
case 8: printf("LXXX"); break;
case 9: printf("XC"); break;
}
n %= 10;
switch(n) {
case 0: break;
case 1: printf("I"); break;
case 2: printf("II"); break;
case 3: printf("III"); break;
case 4: printf("IV"); break;
case 5: printf("V"); break;
case 6: printf("VI"); break;
case 7: printf("VII"); break;
case 8: printf("VIII"); break;
case 9: printf("IX"); break;
}
printf("\n");
}
#define STACKSIZE 1024
int nitems=0;
int stack[STACKSIZE];
int Pop()
{
if (!nitems) {
printf("stack underflow\n");
return 0;
}
return stack[--nitems];
}
void Push(int n)
{
if (nitems == STACKSIZE-1) {
printf("stack overflow\n");
return;
}
stack[nitems++] = n;
}
int PrintTop(void)
{
if (!nitems) {
printf("stack underflow\n");
return -1;
}
else {
PrintRoman(stack[nitems-1]);
return 0;
}
}
int PerformAdd(void)
{
int x,y;
if (nitems < 2) {
printf("stack underflow\n");
return -1;
}
y = Pop();
x = Pop();
Push(x+y);
return 0;
}
int PerformSub(void)
{
int x,y;
if (nitems < 2) {
printf("stack underflow\n");
return -1;
}
y = Pop();
x = Pop();
Push(x-y);
return 0;
}
int PerformMul(void)
{
int x,y;
if (nitems < 2) {
printf("stack underflow\n");
return -1;
}
y = Pop();
x = Pop();
Push(x*y);
return 0;
}
int PerformDiv(void)
{
int x,y;
if (nitems < 2) {
printf("stack underflow\n");
return -1;
}
y = Pop();
x = Pop();
if (y!=0)
Push(x/y);
else {
printf("division by zero exception\n");
Push(x);
}
return 0;
}
main()
{
char line[256];
//freopen("H.in","r",stdin);
// freopen("H.out","w",stdout);
while (fgets(line,sizeof(line),stdin)) {
if (line[0] == '=')
PrintTop();
else if (line[0] == '+')
PerformAdd();
else if (line[0] == '-')
PerformSub();
else if (line[0] == '*')
PerformMul();
else if (line[0] == '/')
PerformDiv();
else {
int n = FromRoman(line);
if (n > 0) {
Push(n);
}
}
}
}
]]>
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狼知道如何用最的代h换取最大的回报
2 l& R/ o) O: F" V7 g" ?+ p; P7 o) j五?nbsp; 同进同退
+ r4 K2 h& D2 u; [- |) tD焉常独自zdQ但狼是最团结的，你不会发现哪只狼在同伴受伤时独自逃走% c9 R" d# D, O. p+ L/ g) o
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九?nbsp; 自由可贵
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' k+ d! U9 {8 o n- }- A狼尊重每个对手，狼在每次d前都会去了解ҎQ而不会去轻视他，所以狼一生的d很少p
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