BNU27937——Soft Kitty——————【扩展前缀和】

Soft Kitty

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format:  %lld      Java class name: Main

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laimao很喜欢这首“BigBang”里的“Soft Kitty”，这首歌的歌词很简单只有6句，"soft kitty, warm kitty, little ball of fur，happy kitty, sleepy kitty, purr purr purr."，她总是唱着玩儿。现在她无聊了决定换一个玩法，你来说出一个数字n，她来唱出第n(1 ≤ n ≤ 10^9)句歌词。注意了她的唱法是，第i次唱这首歌时，每句歌词重复2^(i-1)次。就是像这样，“soft kitty, warm kitty, little ball of fur, happy kitty, sleepy kitty, purr purr purr, soft kitty, soft kitty, warm kitty, warm kitty， little ball of fur, little ball of fur,……”你能帮助她输出第n句歌词么（不包括标点）？

 

Input

多组数据，第一行是一个整数K（0<K<=100），表示数据组数。接下来K行，每行一个数字n，表示你需要输出第n句歌词。

 

Output

输出对应歌词

 

Sample Input

8

1

2

3

4

5

6

7

8

Sample Output

soft kitty

warm kitty

little ball of fur

happy kitty

sleepy kitty

purr purr purr

soft kitty

soft kitty


错误点分析：前缀和计算时出错，数据范围没有考虑好，要用longlong用了int


解题思路：用前缀和的下标来映射唱n首歌。

#include<stdio.h>

#include<string.h>

#include<string>

#include<iostream>

#include<math.h>

#include<algorithm>

using namespace std;

long long f[50];

long long suf[50];

string ss[10];

void work(){



    ss[0]="soft kitty";

    ss[1]="warm kitty";

    ss[2]="little ball of fur";

    ss[3]="happy kitty";

    ss[4]="sleepy kitty";

    ss[5]="purr purr purr";

    f[0]=0;

    suf[0]=0;

    f[1]=6;

    suf[1]=6;

    for(int i=2;i<50;i++){



       f[i]=f[i-1]*2;

       suf[i]=suf[i-1]+f[i];

    }

}

void solve(long long m){



    int i;

    for(i=1;i<50;i++){



        if(suf[i]>=m){



          break;

        }

    }

    m-=suf[i-1];

    long long ti=(long long)pow(2,i-1);

    if(m%ti==0){



        cout<<ss[m/ti-1]<<endl;

    }else{



        cout<<ss[m/ti]<<endl;

    }

}

int main(){



    int K;

    work();

    scanf("%d",&K);

    while(K--){



        long long  n;

        scanf("%lld",&n);

        solve(n);

    }

    return 0;

}