BC12haoj5058&&hdoj5059&&hdoj5560

So easy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 755    Accepted Submission(s): 402

Problem Description

Small W gets two files. There are n integers in each file. Small W wants to know whether these two files are same. So he invites you to write a program to check whether these two files are same. Small W thinks that two files are same when they have the same integer set.
For example file A contains (5,3,7,7),and file B contains (7,5,3,3). They have the same integer set (3,5,7), so they are same.
Another sample file C contains(2,5,2,5), and file D contains (2,5,2,3).
The integer set of C is (2,5),but the integer set of D is (2,3,5),so they are not same.
Now you are expected to write a program to compare two files with size of n.

 

Input

Multi test cases (about 100). Each case contain three lines. The first line contains one integer n represents the size of file. The second line contains n integers a1,a2,a3,…,an - represents the content of the first file. The third line contains n integers b1,b2,b3,…,bn - represents the content of the second file.
Process to the end of file.
1≤n≤100
1≤ai,bi≤1000000000

 

Output

For each case, output "YES" (without quote) if these two files are same, otherwise output "NO" (without quote).

 

Sample Input

31 1 21 2 245 3 7 77 5 3 342 5 2 32 5 2 531 2 31 2 4

 

Sample Output

YESYESNONO

 

Source

BestCoder Round #12

 

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#include<map>using namespace std;const int maxn=110;map<int,int>visa,visb;int numa[maxn];int numb[maxn];int main(){    int t,i,j,k,n,m;    while(scanf("%d",&n)!=EOF){        visa.clear();        visb.clear();        int a,b;        for(i=0;i<n;++i){            scanf("%d",&numa[i]);            visa[numa[i]]++;        }        for(i=0;i<n;++i){            scanf("%d",&numb[i]);            visb[numb[i]]++;        }        for(i=0;i<n;++i){            if(!visa.count(numb[i])||!visb.count(numa[i]))break;        }        if(i<n)            printf("NO\n");        else             printf("YES\n");    }    return 0;}

Help him

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2194    Accepted Submission(s): 445

Problem Description

As you know, when you want to hack someone's program, you must submit your test data. However sometimes you will submit invalid data, so we need a data checker to check your data. Now small W has prepared a problem for BC, but he is too busy to write the data checker. Please help him to write a data check which judges whether the input is an integer ranged from a to b (inclusive).
Note: a string represents a valid integer when it follows below rules.
1. When it represents a non-negative integer, it contains only digits without leading zeros.
2. When it represents a negative integer, it contains exact one negative sign ('-') followed by digits without leading zeros and there are no characters before '-'.
3. Otherwise it is not a valid integer.

 

Input

Multi test cases (about 100), every case occupies two lines, the first line contain a string which represents the input string, then second line contains a and b separated by space. Process to the end of file.

Length of string is no more than 100.
The string may contain any characters other than '\n','\r'.
-1000000000≤a≤b≤1000000000

 

Output

For each case output "YES" (without quote) when the string is an integer ranged from a to b, otherwise output "NO" (without quote).

 

Sample Input

10-100 1001a0-100 100

 

Sample Output

YESNO

 

Source

BestCoder Round #12

 

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>using namespace std;char str[110];char stra[110];char strb[110]; int main(){    int i,j,k;    while(gets(str)){        scanf("%s%s",stra,strb);        getchar();        int len=strlen(str);        bool sign=true;        for(i=0;i<len;++i){            if(str[i]=='-'&&i==0)continue;            if(sign&&str[i]=='0')break;            if(str[i]>='0'&&str[i]<='9');            else break;                        sign=false;        }        if((len==1&&str[0]=='0')){            i=len+1;sign=false;        }        if(i<len||sign){            printf("NO\n");        }        else {            if(str[0]=='-'){                                if(stra[0]=='-'&&strb[0]=='-'){                    if(strcmp(stra+1,str+1)>=0&&strcmp(strb+1,str+1)<=0)                        printf("YES\n");                    else                         printf("NO\n");                }                else if(stra[0]=='-'&&strb[0]!='-'){                    if(strcmp(stra+1,str+1)>=0)                        printf("YES\n");                    else                         printf("NO\n");                }                else {                    printf("NO\n");                }            }            else {                if(stra[0]=='-'&&strb[0]=='-'){                    printf("NO\n");                }                else if(stra[0]=='-'&&strb[0]!='-'){                    if(strcmp(strb,str)>=0)                        printf("YES\n");                    else                         printf("NO\n");                }                else {                    if(strcmp(stra,str)<=0&&strcmp(strb,str)>=0)                        printf("YES\n");                    else                         printf("NO\n");                }            }        }    }    return 0;}

War

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 233    Accepted Submission(s): 76
Special Judge

Problem Description

Long long ago there are two countrys in the universe. Each country haves its own manor in 3-dimension space. Country A's manor occupys x^2+y^2+z^2<=R^2. Country B's manor occupys x^2+y^2<=HR^2 && |z|<=HZ. There may be a war between them. The occurrence of a war have a certain probability. 
We calculate the probability as follow steps.
1. VC=volume of insection manor of A and B.
2. VU=volume of union manor of A and B.
3. probability=VC/VU

 

Input

Multi test cases(about 1000000). Each case contain one line. The first line contains three integers R,HR,HZ. Process to end of file.

[Technical Specification]
0< R,HR,HZ<=100

 

Output

For each case，output the probability of the war which happens between A and B. The answer should accurate to six decimal places.

 

Sample Input

1 1 12 1 1

 

Sample Output

0.6666670.187500

 

Source

BestCoder Round #12

 

题意：求圆柱体与球相交的体积和相并的体积的积

解题思路分情况讨论：球冠的体积计算公式V = πh*h（r-h/3） r为原球的半径h为球冠的高

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#define PI acos(-1.0)using namespace std;const int maxn=10010;int main(){double r,hr,hz;while(scanf("%lf%lf%lf",&r,&hr,&hz)!=EOF){double ans;if(hr>r){if(hz>r){double hc=4.0/3.0*PI*r*r*r;double hu=PI*hr*hr*hz*2.0;ans=hc/hu;}else {double v=PI*(r-hz)*(r-hz)*(r-(r-hz)/3.0)*2.0;double hc=4.0/3.0*PI*r*r*r-v;double hu=PI*hr*hr*hz*2.0+4.0/3.0*PI*r*r*r-hc;ans=hc/hu;}}else {if(hz<=sqrt(r*r-hr*hr)){double hc=PI*hr*hr*hz*2.0;double hu=4.0/3.0*PI*r*r*r;ans=hc/hu;}else if(hz>r){double hh=sqrt(r*r-hr*hr);double hc=PI*hr*hr*hh*2.0+PI*(r-hh)*(r-hh)*(r-(r-hh)/3.0)*2.0;double hu=4.0/3.0*PI*r*r*r+PI*hr*hr*hz*2.0-hc;ans=hc/hu;}else {double hh=sqrt(r*r-hr*hr);double v1=PI*(r-hh)*(r-hh)*(r-(r-hh)/3.0);double v2=PI*(r-hz)*(r-hz)*(r-(r-hz)/3.0);double hc=PI*hr*hr*hh*2.0+(v1-v2)*2.0;double hu=4.0/3.0*PI*r*r*r+PI*hr*hr*hz*2.0-hc;ans=hc/hu;}}printf("%.6lf\n",ans);}return 0;}