Tyvj1053

题目链接

分析：
这道题我觉得比表达式求值还难受

直接模拟
在碰到当前位置的下一个位置是‘-’时
就要考虑展开了

不必展开的所有情况：

1.a-a —-> a-a
-号两边的符号相同
2.a-b，1-2
-号两边的符号紧邻
—-> ab
—-> 12
3.a-4，3-b
两边的符号不同类
—-> a-4
—-> 3-b
4.-号前后也是-
–9 —-> –9

展开

经过重重检验，我们就可以展开了
先按照p3分成两支
按照p1再继续细分

不要忘了判断-号前后的符号既同类又单增

最后如果-号两边的符号是递减的关系
直接复制就好： 4-0 —-> 4-0

tip

当处理完当前-号之后，指针只能向后移动两位，避免以下情况
a-b-c —-> abc
答案数组一定要开够
我在这里开了1w

这里写代码片
#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

int p1,p2,p3,l;
char s[100],ans[10000];

void doit(int bh)
{
    int i,j;
    bool ff=1; 
    if (p3==1)   //正序 
    {
        if (p1==3){   //*
            if (s[bh]>='0'&&s[bh]<='9'&&s[bh+2]<='9'&&s[bh+2]>='0'&&s[bh]<s[bh+2]){
                ans[++l]=s[bh];
                for (j=s[bh]-'0'+1;j<=s[bh+2]-'0'-1;j++)
                    for (i=1;i<=p2;i++)
                        ans[++l]='*';
                ff=0;
            }
            else if (s[bh]>='a'&&s[bh]<='z'&&s[bh+2]<='z'&&s[bh+2]>='a'&&s[bh]<s[bh+2]){
                ans[++l]=s[bh];
                for (j=s[bh]-'a'+1;j<=s[bh+2]-'a'-1;j++)
                    for (i=1;i<=p2;i++)
                        ans[++l]='*';
                ff=0;
            }
        }
        else{
            if (s[bh]>='0'&&s[bh]<='9'&&s[bh+2]<='9'&&s[bh+2]>='0'&&s[bh]<s[bh+2]){
                ans[++l]=s[bh];
                for (j=s[bh]-'0'+1;j<=s[bh+2]-'0'-1;j++)
                    for (i=1;i<=p2;i++)
                        ans[++l]='0'+j;
                ff=0;
            }
            else
            {
                if (p1==1&&s[bh]>='a'&&s[bh]<='z'&&s[bh+2]<='z'&&s[bh+2]>='a'&&s[bh]<s[bh+2]){
                    ans[++l]=s[bh];
                    for (j=s[bh]-'a'+1;j<=s[bh+2]-'a'-1;j++)
                        for (i=1;i<=p2;i++)
                            ans[++l]='a'+j;
                    ff=0;
                }
                else
                if (p1==2&&s[bh]>='a'&&s[bh]<='z'&&s[bh+2]<='z'&&s[bh+2]>='a'&&s[bh]<s[bh+2]){
                    ans[++l]=s[bh];
                    for (j=s[bh]-'a'+1;j<=s[bh+2]-'a'-1;j++)
                        for (i=1;i<=p2;i++)
                            ans[++l]='A'+j;
                    ff=0;
                }
            }
        }   
    }
    else
    {
        if (p1==3){   //*
            if (s[bh]>='0'&&s[bh]<='9'&&s[bh+2]<='9'&&s[bh+2]>='0'&&s[bh]<s[bh+2]){
                ans[++l]=s[bh];
                for (j=s[bh]-'0'+1;j<=s[bh+2]-'0'-1;j++)
                    for (i=1;i<=p2;i++)
                        ans[++l]='*';
                ff=0;
            }
            else if (s[bh]>='a'&&s[bh]<='z'&&s[bh+2]<='z'&&s[bh+2]>='a'&&s[bh]<s[bh+2]){
                ans[++l]=s[bh];
                for (j=s[bh]-'a'+1;j<=s[bh+2]-'a'-1;j++)
                    for (i=1;i<=p2;i++)
                        ans[++l]='*';
                ff=0;
            }
        }
        else{
            if (s[bh]>='0'&&s[bh]<='9'&&s[bh+2]<='9'&&s[bh+2]>='0'&&s[bh]<s[bh+2]){
                ans[++l]=s[bh];
                for (j=s[bh+2]-'0'-1;j>=s[bh]-'0'+1;j--)
                    for (i=1;i<=p2;i++)
                        ans[++l]='0'+j;
                ff=0;
            }
            else
            {
                if (p1==1&&s[bh]>='a'&&s[bh]<='z'&&s[bh+2]<='z'&&s[bh+2]>='a'&&s[bh]<s[bh+2]){
                    ans[++l]=s[bh];
                    for (j=s[bh+2]-'a'-1;j>=s[bh]-'a'+1;j--)
                        for (i=1;i<=p2;i++)
                            ans[++l]='a'+j;
                    ff=0;
                }
                else
                if (p1==2&&s[bh]>='a'&&s[bh]<='z'&&s[bh+2]<='z'&&s[bh+2]>='a'&&s[bh]<s[bh+2]){
                    ans[++l]=s[bh];
                    for (j=s[bh+2]-'a'-1;j>=s[bh]-'a'+1;j--)
                        for (i=1;i<=p2;i++)
                            ans[++l]='A'+j;
                    ff=0;
                }
            }
        }   
    }
    if (ff) ans[++l]=s[bh],ans[++l]=s[bh+1];
}

int main()
{
    scanf("%d%d%d",&p1,&p2,&p3);
    scanf("%s",&s);
    int i=0,len=strlen(s);
    l=-1;
    while (i<len)
    {
        if (s[i+1]=='-')  //要展开了 
        {
            if (s[i]==s[i+2]){
                ans[++l]=s[i]; ans[++l]=s[i+1]; i=i+2;
            }
            else if ((int)s[i]==(int)s[i+2]-1){
                ans[++l]=s[i]; i=i+2;
            }
            else if ((s[i]>='0'&&s[i]<='9'&&s[i+2]>='a'&&s[i+2]<='z')||(s[i+2]>='0'&&s[i+2]<='9'&&s[i]>='a'&&s[i]<='z')){
                ans[++l]=s[i]; ans[++l]=s[i+1]; i+=2;
            }   
            else if (s[i]==s[i+1]){
                ans[++l]=s[i]; ans[++l]=s[i+1]; i+=2;
            }
            else if (s[i+1]==s[i+2]){
                ans[++l]=s[i]; ans[++l]=s[i+1]; ans[++l]=s[i+2]; i+=3;
            }
            else doit(i),i+=2;
        }
        else
        {
            ans[++l]=s[i]; i++;
        }
    }
    printf("%s",ans);
    return 0;
}