注意这里的优化:我们先将所有操作读进内存,从后往前计算出哪些操作是必要的,然后仅仅执行必要的操作就行了
至于哪些操作的是必要的呢,我们需要从后往前维护当前哪些变量的值是必要的,
首先,程序结束的时候,显然所有变量的值都是必要的。
每当遇到一个赋值语句的时候,被覆盖的变量的值之前就是不必要的了。
每当遇到一个计算语句的时候,如果修改的变量是必要的,那么显然用来作为运算的变量也就变为必要的了
Accepted | 3546 | 15MS | 1988K | 1633 B | C++ |


#include
<
iostream
>
#include < cmath >
#include < cstring >
using namespace std;
int n = 0 ;
char s[ 300003 ][ 5 ];
bool useful[ 300003 ];
bool u[ 10 ];
const __int64 base = 1000000000 ;
struct integer{
__int64 a[ 600 ];
int length;
integer(){
memset(a, 0 , sizeof (a));
a[ 0 ] = 1 ; length = 0 ;
}
void operator += (integer x)
{
__int64 t = 0 ;
if (x.length > length) length = x.length;
length += 5 ;
for ( int i = 0 ; i < length;i ++ )
{
a[i] += (x.a[i] + t);
if (a[i] >= base ) a[i] -= base , t = 1 ;
else t = 0 ;
}
length = 599 ;
while (a[length] == 0 && length > 0 ) length -- ;
}
void operator *= (integer x)
{
__int64 res[ 600 ],t;
memset(res, 0 , sizeof (res));
for ( int i = 0 ;i <= length;i ++ )
{
t = 0 ;
for ( int j = 0 ; ! (j > x.length && t == 0 );j ++ )
{
res[i + j] += (a[i] * x.a[j] + t);
t = res[i + j] / base ;
res[i + j] %= base ;
}
}
memcpy(a,res, sizeof (res));
length = 599 ;
while (a[length] == 0 && length > 0 ) length -- ;
}
void output()
{
int i = length;
printf( " %I64d " ,a[i]); i -- ;
for ( ;i >= 0 ;i -- ) printf( " %09I64d " ,a[i]);
puts( "" );
}
}ans[ 10 ];
int main()
{
int i;
while (gets(s[n])) n ++ ;
for (i = 0 ;i < 10 ;i ++ )
u[i] = true ;
for (i = n - 1 ;i >= 0 ;i -- )
{
useful[i] = u[s[i][ 0 ] - ' A ' ];
if (s[i][ 3 ] == 0 )
u[s[i][ 0 ] - ' A ' ] = false ;
else {
if (u[s[i][ 0 ] - ' A ' ] == true )
u[s[i][ 3 ] - ' A ' ] = true ;
}
}
for (i = 0 ;i < n;i ++ )
if (useful[i])
{
if (s[i][ 3 ] == 0 )
ans[s[i][ 0 ] - ' A ' ] = ans[s[i][ 2 ] - ' A ' ];
else if (s[i][ 1 ] == ' + ' )
ans[s[i][ 0 ] - ' A ' ] += ans[s[i][ 3 ] - ' A ' ];
else if (s[i][ 1 ] == ' * ' )
ans[s[i][ 0 ] - ' A ' ] *= ans[s[i][ 3 ] - ' A ' ];
}
for (i = 0 ;i < 10 ;i ++ )
ans[i].output();
return 0 ;
}
#include < cmath >
#include < cstring >
using namespace std;
int n = 0 ;
char s[ 300003 ][ 5 ];
bool useful[ 300003 ];
bool u[ 10 ];
const __int64 base = 1000000000 ;
struct integer{
__int64 a[ 600 ];
int length;
integer(){
memset(a, 0 , sizeof (a));
a[ 0 ] = 1 ; length = 0 ;
}
void operator += (integer x)
{
__int64 t = 0 ;
if (x.length > length) length = x.length;
length += 5 ;
for ( int i = 0 ; i < length;i ++ )
{
a[i] += (x.a[i] + t);
if (a[i] >= base ) a[i] -= base , t = 1 ;
else t = 0 ;
}
length = 599 ;
while (a[length] == 0 && length > 0 ) length -- ;
}
void operator *= (integer x)
{
__int64 res[ 600 ],t;
memset(res, 0 , sizeof (res));
for ( int i = 0 ;i <= length;i ++ )
{
t = 0 ;
for ( int j = 0 ; ! (j > x.length && t == 0 );j ++ )
{
res[i + j] += (a[i] * x.a[j] + t);
t = res[i + j] / base ;
res[i + j] %= base ;
}
}
memcpy(a,res, sizeof (res));
length = 599 ;
while (a[length] == 0 && length > 0 ) length -- ;
}
void output()
{
int i = length;
printf( " %I64d " ,a[i]); i -- ;
for ( ;i >= 0 ;i -- ) printf( " %09I64d " ,a[i]);
puts( "" );
}
}ans[ 10 ];
int main()
{
int i;
while (gets(s[n])) n ++ ;
for (i = 0 ;i < 10 ;i ++ )
u[i] = true ;
for (i = n - 1 ;i >= 0 ;i -- )
{
useful[i] = u[s[i][ 0 ] - ' A ' ];
if (s[i][ 3 ] == 0 )
u[s[i][ 0 ] - ' A ' ] = false ;
else {
if (u[s[i][ 0 ] - ' A ' ] == true )
u[s[i][ 3 ] - ' A ' ] = true ;
}
}
for (i = 0 ;i < n;i ++ )
if (useful[i])
{
if (s[i][ 3 ] == 0 )
ans[s[i][ 0 ] - ' A ' ] = ans[s[i][ 2 ] - ' A ' ];
else if (s[i][ 1 ] == ' + ' )
ans[s[i][ 0 ] - ' A ' ] += ans[s[i][ 3 ] - ' A ' ];
else if (s[i][ 1 ] == ' * ' )
ans[s[i][ 0 ] - ' A ' ] *= ans[s[i][ 3 ] - ' A ' ];
}
for (i = 0 ;i < 10 ;i ++ )
ans[i].output();
return 0 ;
}
这一题用JAVA处理大数很easy,但是不会JAVA程序(表示很弱)……
附上大牛编写的JAVA版代码,供ym(仰慕)。
Accepted | 3546 | 1375MS | 5268K | 790 B | Java |
import
java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math. * ;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
int n = 10 ;
BigInteger[] mem = new BigInteger[n];
for ( int i = 0 ;i < n; ++ i){
mem[i] = BigInteger.ONE;
}
while ( true ){
String s = br.readLine();
if (s == null ) break ;
if (s.length() == 3 ){
mem[s.charAt( 0 ) - ' A ' ] = mem[s.charAt( 2 ) - ' A ' ];
} else {
int n1 = s.charAt( 0 ) - ' A ' ,n2 = s.charAt( 3 ) - ' A ' ;
if (s.charAt( 1 ) == ' * ' ){
mem[n1] = mem[n1].multiply(mem[n2]);
} else {
mem[n1] = mem[n1].add(mem[n2]);
}
}
}
for ( int i = 0 ;i < n; ++ i){
System.out.println(mem[i]);
}
}
}
import java.io.IOException;
import java.io.InputStreamReader;
import java.math. * ;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
int n = 10 ;
BigInteger[] mem = new BigInteger[n];
for ( int i = 0 ;i < n; ++ i){
mem[i] = BigInteger.ONE;
}
while ( true ){
String s = br.readLine();
if (s == null ) break ;
if (s.length() == 3 ){
mem[s.charAt( 0 ) - ' A ' ] = mem[s.charAt( 2 ) - ' A ' ];
} else {
int n1 = s.charAt( 0 ) - ' A ' ,n2 = s.charAt( 3 ) - ' A ' ;
if (s.charAt( 1 ) == ' * ' ){
mem[n1] = mem[n1].multiply(mem[n2]);
} else {
mem[n1] = mem[n1].add(mem[n2]);
}
}
}
for ( int i = 0 ;i < n; ++ i){
System.out.println(mem[i]);
}
}
}