zoukankan      html  css  js  c++  java
  • Ural 2045. Richness of words 打表找规律

    2045. Richness of words

    题目连接:

    http://acm.timus.ru/problem.aspx?space=1&num=2045

    Description

    For each integer i from 1 to n, you must print a string si of length n consisting of lowercase Latin letters. The string si must contain exactly i distinct palindrome substrings. Two substrings are considered distinct if they are different as strings.

    Input

    The input contains one integer n (1 ≤ n ≤ 2000).

    Output

    You must print n lines. If for some i, the answer exists, print it in the form “i : si” where si is one of possible strings. Otherwise, print “i : NO”.

    Sample Input

    4

    Sample Output

    1 : NO
    2 : NO
    3 : abca
    4 : bbca

    Hint

    题意

    给你n,让你构造长度为n的字符串,使得这个字符串本质不同的回文串恰好i个

    无解输出No

    题解:

    打表找规律,发现是abc->aabc->aaabc->aaaabc这样的循环就好了

    卡了输入和输出

    代码

    #include<bits/stdc++.h>
    using namespace std;
    namespace fastIO{
        #define BUF_SIZE 100000
        #define OUT_SIZE 100000
        #define ll long long
        //fread->read
        bool IOerror=0;
        inline char nc(){
            static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
            if (p1==pend){
                p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
                if (pend==p1){IOerror=1;return -1;}
                //{printf("IO error!
    ");system("pause");for (;;);exit(0);}
            }
            return *p1++;
        }
        inline bool blank(char ch){return ch==' '||ch=='
    '||ch=='
    '||ch=='	';}
        inline void read(int &x){
            bool sign=0; char ch=nc(); x=0;
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            if (ch=='-')sign=1,ch=nc();
            for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
            if (sign)x=-x;
        }
        inline void read(ll &x){
            bool sign=0; char ch=nc(); x=0;
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            if (ch=='-')sign=1,ch=nc();
            for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
            if (sign)x=-x;
        }
        inline void read(double &x){
            bool sign=0; char ch=nc(); x=0;
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            if (ch=='-')sign=1,ch=nc();
            for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
            if (ch=='.'){
                double tmp=1; ch=nc();
                for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
            }
            if (sign)x=-x;
        }
        inline void read(char *s){
            char ch=nc();
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
            *s=0;
        }
        inline void read(char &c){
            for (c=nc();blank(c);c=nc());
            if (IOerror){c=-1;return;}
        }
        //getchar->read
        inline void read1(int &x){
            char ch;int bo=0;x=0;
            for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
            for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
            if (bo)x=-x;
        }
        inline void read1(ll &x){
            char ch;int bo=0;x=0;
            for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
            for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
            if (bo)x=-x;
        }
        inline void read1(double &x){
            char ch;int bo=0;x=0;
            for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
            for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
            if (ch=='.'){
                double tmp=1;
                for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar());
            }
            if (bo)x=-x;
        }
        inline void read1(char *s){
            char ch=getchar();
            for (;blank(ch);ch=getchar());
            for (;!blank(ch);ch=getchar())*s++=ch;
            *s=0;
        }
        inline void read1(char &c){for (c=getchar();blank(c);c=getchar());}
        //scanf->read
        inline void read2(int &x){scanf("%d",&x);}
        inline void read2(ll &x){
            #ifdef _WIN32
                scanf("%I64d",&x);
            #else
            #ifdef __linux
                scanf("%lld",&x);
            #else
                puts("error:can't recognize the system!");
            #endif
            #endif
        }
        inline void read2(double &x){scanf("%lf",&x);}
        inline void read2(char *s){scanf("%s",s);}
        inline void read2(char &c){scanf(" %c",&c);}
        inline void readln2(char *s){gets(s);}
        //fwrite->write
        struct Ostream_fwrite{
            char *buf,*p1,*pend;
            Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
            void out(char ch){
                if (p1==pend){
                    fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
                }
                *p1++=ch;
            }
            void print(int x){
                static char s[15],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1);
            }
            void println(int x){
                static char s[15],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1); out('
    ');
            }
            void print(ll x){
                static char s[25],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1);
            }
            void println(ll x){
                static char s[25],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1); out('
    ');
            }
            void print(double x,int y){
                static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,
                    1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,
                    100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};
                if (x<-1e-12)out('-'),x=-x;x*=mul[y];
                ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;
                ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);
                if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}
            }
            void println(double x,int y){print(x,y);out('
    ');}
            void print(char *s){while (*s)out(*s++);}
            void println(char *s){while (*s)out(*s++);out('
    ');}
            void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
            ~Ostream_fwrite(){flush();}
        }Ostream;
        inline void print(int x){Ostream.print(x);}
        inline void println(int x){Ostream.println(x);}
        inline void print(char x){Ostream.out(x);}
        inline void println(char x){Ostream.out(x);Ostream.out('
    ');}
        inline void print(ll x){Ostream.print(x);}
        inline void println(ll x){Ostream.println(x);}
        inline void print(double x,int y){Ostream.print(x,y);}
        inline void println(double x,int y){Ostream.println(x,y);}
        inline void print(char *s){Ostream.print(s);}
        inline void println(char *s){Ostream.println(s);}
        inline void println(){Ostream.out('
    ');}
        inline void flush(){Ostream.flush();}
        //puts->write
        char Out[OUT_SIZE],*o=Out;
        inline void print1(int x){
            static char buf[15];
            char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
            while(x)*p1++=x%10+'0',x/=10;
            while(p1--!=buf)*o++=*p1;
        }
        inline void println1(int x){print1(x);*o++='
    ';}
        inline void print1(ll x){
            static char buf[25];
            char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
            while(x)*p1++=x%10+'0',x/=10;
            while(p1--!=buf)*o++=*p1;
        }
        inline void println1(ll x){print1(x);*o++='
    ';}
        inline void print1(char c){*o++=c;}
        inline void println1(char c){*o++=c;*o++='
    ';}
        inline void print1(char *s){while (*s)*o++=*s++;}
        inline void println1(char *s){print1(s);*o++='
    ';}
        inline void println1(){*o++='
    ';}
        inline void flush1(){if (o!=Out){if (*(o-1)=='
    ')*--o=0;puts(Out);}}
        struct puts_write{
            ~puts_write(){flush1();}
        }_puts;
        inline void print2(int x){printf("%d",x);}
        inline void println2(int x){printf("%d
    ",x);}
        inline void print2(char x){printf("%c",x);}
        inline void println2(char x){printf("%c
    ",x);}
        inline void print2(ll x){
            #ifdef _WIN32
                printf("%I64d",x);
            #else
            #ifdef __linux
                printf("%lld",x);
            #else
                puts("error:can't recognize the system!");
            #endif
            #endif
        }
        inline void println2(ll x){print2(x);printf("
    ");}
        inline void println2(){printf("
    ");}
        #undef ll
        #undef OUT_SIZE
        #undef BUF_SIZE
    };
    using namespace fastIO;
    
    char output[2005];
    int main(){
        int n;
        scanf("%d",&n);
        if(n==1){
            printf("1 : a
    ");
            return 0;
        }
        if(n==2){
            printf("1 : NO
    ");
            printf("2 : aa
    ");
            return 0;
        }
        printf("1 : NO
    ");
        printf("2 : NO
    ");
        for(int i=3;i<n;i++){
            printf("%d : ",i);
            for(int j=0;j<n;j++){
                if(j%i<i-2)putchar('a');
                else if(j%i==i-2)putchar('b');
                else putchar('c');
            }
            puts("");
        }
        printf("%d : ",n);
        for(int i=0;i<n;i++)
            putchar('a');
        puts("");
    }
  • 相关阅读:
    C# 建立快捷方式
    ae中gp执行 Error HRESULT E_FAIL has been returned from a call to a COM component错误
    AE编辑点要素编辑
    噱头还是革命 云计算“泡沫”五年后改变世界? 狼人:
    分析:英特尔收购McAfee的三大意义 狼人:
    云安全:防护“工具”还是攻击“利器” 狼人:
    热点:安全问题是否能将DNS推入云服务 狼人:
    迈克菲收购tenCube 打造新一代移动安全平台 狼人:
    戴尔推免费浏览器安全工具 可隔离恶意软件 狼人:
    黑帽大会:HTTPS和SSL协议存在安全漏洞 狼人:
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5742576.html
Copyright © 2011-2022 走看看