The 2014 ACM-ICPC Asia Mudanjiang Regional Contest - K Known Notation
2014-10-12 16:29
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*号能运算时必须栈里至少有两个数字,如果当前不能就将这个*和最后一个数字交换(*号当然越靠后越好)
数字只需要考虑够不够的情况,n个*至少n+1个数字,因为数字可以任意拆,如果多了也无所谓,而且如果最后一个数字后面没有*那么之前肯定是可以出现*号换到后面的情况
#include<stdio.h>
#include<string.h>
#define MAXN 1005
char s[MAXN];
int len;
void sw(int x)
{
for(int i=len-1;i>=0;i--)
{
if('1'<=s[i]&&s[i]<='9')
{
char t=s[i];
s[i]=s[x];
s[x]=t;
return ;
}
}
}
int main()
{
int cas;
scanf("%d",&cas);
while(cas--)
{
scanf("%s",s);
len=(int)strlen(s);
int cnt1=0,cnt2=0;
for(int i=0;i<len;i++)
if(s[i]=='*') cnt2++;
else cnt1++;
int st=0;//模拟栈
int ans=0;
if(cnt1<(cnt2+1)) ans=cnt2+1-cnt1,st+=ans;
for(int i=0;i<len;i++)
{
if('1'<=s[i]&&s[i]<='9') st++;
else
{
if(st>=2) st--;
else
{
ans++;
sw(i);
i--;
}
}
}
printf("%d\n",ans);
}
return 0;
}
数字只需要考虑够不够的情况,n个*至少n+1个数字,因为数字可以任意拆,如果多了也无所谓,而且如果最后一个数字后面没有*那么之前肯定是可以出现*号换到后面的情况
#include<stdio.h>
#include<string.h>
#define MAXN 1005
char s[MAXN];
int len;
void sw(int x)
{
for(int i=len-1;i>=0;i--)
{
if('1'<=s[i]&&s[i]<='9')
{
char t=s[i];
s[i]=s[x];
s[x]=t;
return ;
}
}
}
int main()
{
int cas;
scanf("%d",&cas);
while(cas--)
{
scanf("%s",s);
len=(int)strlen(s);
int cnt1=0,cnt2=0;
for(int i=0;i<len;i++)
if(s[i]=='*') cnt2++;
else cnt1++;
int st=0;//模拟栈
int ans=0;
if(cnt1<(cnt2+1)) ans=cnt2+1-cnt1,st+=ans;
for(int i=0;i<len;i++)
{
if('1'<=s[i]&&s[i]<='9') st++;
else
{
if(st>=2) st--;
else
{
ans++;
sw(i);
i--;
}
}
}
printf("%d\n",ans);
}
return 0;
}
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