hdoj 5592 ZYB's Biology 【线段树】 hdoj 5590 ZYB's Biology 【字符串水题】

ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 340    Accepted Submission(s): 281

Problem Description

After getting 600 scores in NOIPZYB(ZJ−267) begins to work with biological questions.Now he give you a simple biological questions:
he gives you a DNA sequence and a RNA sequence,then he asks you whether the DNA sequence and the RNA sequence are
matched.

The DNA sequence is a string consisted of A,C,G,T;TheRNA sequence is a string consisted of A,C,G,U.

DNA sequence and RNA sequence are matched if and only if A matches U,T matches A,C matches G,G matches C on each position.

 

Input

In the first line there is the testcase T.

For each teatcase:

In the first line there is one number N.

In the next line there is a string of length N,describe the DNA sequence.

In the third line there is a string of length N,describe the RNA sequence.

1≤T≤10,1≤N≤100

 

Output

For each testcase,print YES or NO,describe whether the two arrays are matched.

 

Sample Input

24ACGTUGCA4ACGTACGU

 

Sample Output

YESNO

 

Source

BestCoder Round #65 

恩，题目大意就是说，给出一个序列到相应位置的逆序数，求原序列。本来想的是求当前位置的数前有几个数比它大就把它插到从后面数几个数之前，这个用for循环写的超时。。。然后现在是找到当前位置的数前有x个数比它大，那么它就是第x+1大的数，倒着插。（看的雨神代码，好久才懂）

<span style="font-family:Comic Sans MS;font-size:18px;">#include<cstdio>#include<cstring>#include<algorithm>#define maxn 50050using namespace std;int n,a[maxn],b[maxn];struct Node{    int l,r,s;};Node node[maxn<<2];void pushup(int o){    node[o].s=node[o<<1].s+node[o<<1|1].s;}void build(int o,int l,int r){    node[o].l=l;    node[o].r=r;    node[o].s=1;    if(l==r)        return ;    int mid=(l+r)>>1;    build(o<<1,l,mid);    build(o<<1|1,mid+1,r);    pushup(o);}void update(int o,int aim){    if(node[o].l==node[o].r)    {        node[o].s=0;        return ;    }    int mid=(node[o].l+node[o].r)>>1;    if(aim<=mid)        update(o<<1,aim);    else        update(o<<1|1,aim);    pushup(o);}int query(int o,int aim){    if(node[o].l==node[o].r)        return node[o].l;    if(node[o<<1].s>=aim)        return query(o<<1,aim);    else        return query(o<<1|1,aim-node[o<<1].s);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;++i)            scanf("%d",&a[i]);        a[0]=0;        build(1,1,n);        for(int i=n;i>=1;i--)        {            int tem=a[i]-a[i-1];            int p=query(1,tem+1);            b[i]=n-p+1;            update(1,p);        }        for(int i=1;i<=n;++i)        {            if(i==1)                printf("%d",b[i]);            else                printf(" %d",b[i]);        }        printf("\n");    }    return 0;}</span>

恩，打BC时的代码，本来以为做对了，又被hack掉了，超时。。。。（还是要用上面线段树做，这个是从前面开始的，其实思想都差不多）

<span style="font-family:Comic Sans MS;font-size:18px;background-color: rgb(192, 192, 192);">#include <iostream>#include<cstdio>#include<cstring>#define maxn 50050using namespace std;int a[maxn],b[maxn];void in(int x,int pos){    for(int i=x;i>x-pos;--i)    {        b[i]=b[i-1];    }    b[x-pos]=x;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;++i)            scanf("%d",&a[i]);        b[1]=1;        int s=1;        for(int i=2;i<=n;++i)        {            if(!a[i])                b[i]=i;            else            {                int tem=a[i]-a[i-1];                in(i,tem);            }        }        for(int i=1;i<=n;++i)            a[b[i]]=i;        for(int i=1;i<=n;++i)        {            if(i==1)                printf("%d",a[i]);            else                printf(" %d",a[i]);        }        printf("\n");    }    return 0;}</span>

ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 340    Accepted Submission(s): 281

Problem Description

After getting 600 scores in NOIPZYB(ZJ−267) begins to work with biological questions.Now he give you a simple biological questions:
he gives you a DNA sequence and a RNA sequence,then he asks you whether the DNA sequence and the RNA sequence are
matched.

The DNA sequence is a string consisted of A,C,G,T;TheRNA sequence is a string consisted of A,C,G,U.

DNA sequence and RNA sequence are matched if and only if A matches U,T matches A,C matches G,G matches C on each position.

 

Input

In the first line there is the testcase T.

For each teatcase:

In the first line there is one number N.

In the next line there is a string of length N,describe the DNA sequence.

In the third line there is a string of length N,describe the RNA sequence.

1≤T≤10,1≤N≤100

 

Output

For each testcase,print YES or NO,describe whether the two arrays are matched.

 

Sample Input

24ACGTUGCA4ACGTACGU

 

Sample Output

YESNO

 

Source

BestCoder Round #65 

恩，字符串水题

<span style="font-family:Comic Sans MS;font-size:18px;background-color: rgb(204, 204, 204);">#include <iostream>#include<cstdio>#include<cstring>#define maxn 110using namespace std;char dna[maxn],rna[maxn];int judge(int x){    if(dna[x]=='A'&&rna[x]=='U')        return 1;    if(dna[x]=='C'&&rna[x]=='G')        return 1;    if(dna[x]=='T'&&rna[x]=='A')        return 1;    if(dna[x]=='G'&&rna[x]=='C')        return 1;    return 0;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        getchar();        for(int i=0;i<n;++i)            scanf("%c",&dna[i]);        getchar();        for(int i=0;i<n;++i)            scanf("%c",&rna[i]);        int flag=0;        for(int i=0;i<n;++i)        {            if(!judge(i))            {                flag=1;                break;            }        }        if(flag)            printf("NO\n");        else            printf("YES\n");    }    return 0;}</span>