A.GPA(HDU4802):
纯属进入状态用,给你一些字符串对应的权重,求加权平均,如果是N,P不计入统计
GPA
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1193 Accepted Submission(s): 743
Problem Description
In college, a student may take several courses. for each course i, he earns a certain credit (ci), and a mark ranging from A to F, which is comparable to a score (si), according to the following conversion table
The GPA is the weighted average score of all courses one student may take, if we treat the credit as the weight. In other words,
An additional treatment is taken for special cases. Some courses are based on “Pass/Not pass” policy, where stude nts earn a mark “P” for “Pass” and a mark “N” for “Not pass”. Such courses are not supposed to be considered in computation. These special courses must be ignored for computing the correct GPA.
Specially, if a student’s credit in GPA computation is 0, his/her GPA will be “0.00”.
Input
There are several test cases, please process till EOF.
Each test case starts with a line containing one integer N (1 <= N <= 1000), the number of courses. Then follows N lines, each consisting the credit and the mark of one course. Credit is a positive integer and less than 10.
Output
For each test case, print the GPA (rounded to two decimal places) as the answer.
Sample Input
5
2 B
3 D-
2 P
1 F
3 A
2
2 P
2 N
6
4 A
3 A
3 A
4 A
3 A
3 A
Sample Output
2.33
0.00
4.00
Hint
For the first test case:
GPA =(3.0 * 2 + 1.0 * 3 + 0.0 * 1 + 4.0 * 3)/(2 + 3 + 1 + 3) = 2.33
For the second test case: because credit in GPA computation is 0(P/N in additional treatment), so his/her GPA is “0.00”.
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int main()
{
int n,nn,num;
double res,mid;
char cc[5];
while(scanf("%d",&n)!=EOF){
res=0;
num=0;
for(int k=0;k<n;k++){
mid=0;
scanf("%d%s",&nn,cc);
if(cc[0]==‘P‘||cc[0]==‘N‘) continue;
if(cc[0]==‘A‘){
if(cc[1]==‘\0‘) mid+=4.0;
else if(cc[1]==‘-‘) mid+=3.7;
}
else if(cc[0]==‘B‘){
if(cc[1]==‘\0‘) mid+=3.0;
else if(cc[1]==‘+‘) mid+=3.3;
else if(cc[1]==‘-‘) mid+=2.7;
}
else if(cc[0]==‘C‘){
if(cc[1]==‘\0‘) mid+=2.0;
else if(cc[1]==‘+‘) mid+=2.3;
else if(cc[1]==‘-‘) mid+=1.7;
}
else if(cc[0]==‘D‘){
if(cc[1]==‘\0‘) mid+=1.3;
else if(cc[1]==‘-‘) mid+=1.0;
}
num+=nn;
res+=mid*nn;
}
if(num==0) printf("0.00\n");
else printf("%.2f\n",res/num);
}
return 0;
}
B.Poor Warehouse Keeper(HDU4803)
Poor Warehouse Keeper
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1089 Accepted Submission(s): 312
Problem Description
Jenny is a warehouse keeper. He writes down the entry records everyday. The record is shown on a screen, as follow:
There are only two buttons on the screen. Pressing the button in the first line once increases the number on the first line by 1. The cost per unit remains untouched. For the screen above, after the button in the first line is pressed, the screen will be:
The exact total price is 7.5, but on the screen, only the integral part 7 is shown.
Pressing the button in the second line once increases the number on the second line by 1. The number in the first line remains untouched. For the screen above, after the button in the second line is pressed, the screen will be:
Remember the exact total price is 8.5, but on the screen, only the integral part 8 is shown.
A new record will be like the following:
At that moment, the total price is exact 1.0.
Jenny expects a final screen in form of:
Where x and y are previously given.
What’s the minimal number of pressing of buttons Jenny needs to achieve his goal?
Input
There are several (about 50, 000) test cases, please process till EOF.
Each test case contains one line with two integers x(1 <= x <= 10) and y(1 <= y <= 109) separated by a single space - the expected number shown on the screen in the end.
Output
For each test case, print the minimal number of pressing of the buttons, or “-1”(without quotes) if there’s no way to achieve his goal.
Sample Input
1 1
3 8
9 31
Sample Output
0
5
11
Hint
For the second test case, one way to achieve is:
(1, 1) -> (1, 2) -> (2, 4) -> (2, 5) -> (3, 7.5) -> (3, 8.5)
题意:有个电子屏 上x,下y,上下各有按钮,上面的按钮保持分子分母比例使x+1,(也即x/y=(x+1)/(y+deltay)),下面的直接y+1,小数不显示但是不约去
赛时总结:打乱了smilewsw的节奏.....这题没做出来有三点
1.smile笔算算错了使我以为是下界没找对
2. 应该稍微比下界少一点就行了,这里的一点不是deltay为整数的时候就直接-1
3. double的运算应该放到一起去,不能先得到(y+deltay)换算成整数后再减去y,而是应该在得到整数前减去y,忽略了y可能是个小数这一点
解题思路:贪心,肯定应该介于x/(y+1),x/(y-1)之间,又因为比例越小,那么按上面的键y增的越多,所以不用管x/(y-1),尽量逼近但是又不超过x/(y+1)即可,所以设下限为x/(y+1-eps),
贪心证明在于这个比例只能被扩大不能缩小
#include<cstdio>
using namespace std;
int x,y;
int main()
{
while(scanf("%d%d",&x,&y)==2){
if(x>y){puts("-1");continue;}
if(x==1||x==y){printf("%d\n",y-1);continue;}
double sub,dow;
sub=x;
dow=y+1-0.01;
double ty=1;//用来放置过程中的y,过程中的x为i
int ans=x-1;//肯定要增减x-1次,用于上方的键
for(int i=1;i<=x;i++){
int deltay=(int)((i*dow)/sub-ty);//按下方键的次数
ans+=deltay;
ty+=deltay;
ty=ty*(i+1)/i;//按了上方的键
}
printf("%d\n",ans);
}
return 0;
}
C. Campus Design(HDU4804):
Campus Design
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 406 Accepted Submission(s): 201
Problem Description
Nanjing University of Science and Technology is celebrating its 60th anniversary. In order to make room for student activities, to make the university a more pleasant place for learning, and to beautify the campus, the college administrator decided to start construction on an open space.
The designers measured the open space and come to a conclusion that the open space is a rectangle with a length of n meters and a width of m meters. Then they split the open space into n x m squares. To make it more beautiful, the designer decides to cover the open space with 1 x 1 bricks and 1 x 2 bricks, according to the following rules:
1. All the bricks can be placed horizontally or vertically
2. The vertexes of the bricks should be placed on integer lattice points
3. The number of 1 x 1 bricks shouldn’t be less than C or more than D. The number of 1 x 2 bricks is unlimited.
4. Some squares have a flowerbed on it, so it should not be covered by any brick. (We use 0 to represent a square with flowerbet and 1 to represent other squares)
Now the designers want to know how many ways are there to cover the open space, meeting the above requirements.
Input
There are several test cases, please process till EOF.
Each test case starts with a line containing four integers N(1 <= N <= 100), M(1 <= M <= 10), C, D(1 <= C <= D <= 20). Then following N lines, each being a string with the length of M. The string consists of ‘0’ and ‘1’ only, where ‘0’ means the square should not be covered by any brick, and ‘1’ otherwise.
Output
Please print one line per test case. Each line should contain an integers representing the answer to the problem (mod 109 + 7).
Sample Input
1 1 0 0
1
1 1 1 2
0
1 1 1 2
1
1 2 1 2
11
1 2 0 2
01
1 2 0 2
11
2 2 0 0
10
10
2 2 0 0
01
10
2 2 0 0
11
11
4 5 3 5
11111
11011
10101
11111
Sample Output
0
0
1
1
1
2
1
0
2
954
题意:轮廓线DP
总结:不要无谓跟榜啊....I题缺少了关键性的,C和网络流却可以一发出的...微笑大神beautiful!明白轮廓线dp了
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>
#define MOD 1000000007
using namespace std;
long long dp[2][1<<11][21];
char a[200][50];
int main()
{
int n,m,c,d,i,j,used,k,cnt;
while(scanf("%d %d %d %d",&n,&m,&c,&d)!=EOF)
{
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++)
scanf("%s",a[i]);
dp[0][0][0]=1;
cnt=0;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
memset(dp[1-cnt%2],0,sizeof(dp[1-cnt%2]));
for(used=0;used<1<<m;used++)
{
for(k=0;k<=d;k++)
{
if((used>>j&1)||a[i][j]==‘0‘)
dp[1-cnt%2][used&~(1<<j)][k]=(dp[1-cnt%2][used&~(1<<j)][k]+dp[cnt%2][used][k])%MOD;
else
{
if(k<d)
dp[1-cnt%2][used&~(1<<j)][k+1]=(dp[1-cnt%2][used&~(1<<j)][k+1]+dp[cnt%2][used][k])%MOD;
if(j+1<m&&!(used>>(j+1)&1)&&a[i][j+1]==‘1‘)
dp[1-cnt%2][used|1<<(j+1)][k]=(dp[1-cnt%2][used|1<<(j+1)][k]+dp[cnt%2][used][k])%MOD;
if(i+1<n&&a[i+1][j]==‘1‘)
dp[1-cnt%2][used|1<<j][k]=(dp[1-cnt%2][used|1<<j][k]+dp[cnt%2][used][k])%MOD;
}
}
}
cnt++;
}
}
long long sum=0;
for(i=c;i<=d;i++)
{
sum=(sum+dp[cnt%2][0][i])%MOD;
}
printf("%I64d\n",sum);
}
return 0;
}
D. Shoot(HDU4805):
Shoot
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 315 Accepted Submission(s): 50
Special Judge
Problem Description
At the year of 8192, the war between Evil Army and Galaxy Army broken out. Unfortunately, Evil Army had conquered half the galaxy in just one year. To prevent the situation of the war from getting worse, Levi, the general of Galaxy Elite Army, was ordered by his superior to attack the enemy’s power bases.
Levi was born with the remarkable ability of counter-surveillance, it was just a piece of cake for him to reach Evil Army’s power bases. Each power base can be represented as a triangle in 3D-Cartesian coordinate system. The only weapon Levi had was a laser cannon which can shoot in both two directions simultaneously. To avoid being caught by enemy, Levi can only place the laser cannon somewhere on a segment from S to T. Unfortunately, there was something wrong with the laser cannon, Levi can’t adjust its shooting angle, so the shooting direction was fixed.
Since Levi didn’t have any time to find the best place to shoot the laser, he decided to select a point on the segment randomly to place the cannon. If the laser touched the base(even the boundary), the base will be destroyed. Your task is to calculate the expected number of the destroyed bases in just one shoot.
It is recommended to see the sample input to understand the problem statement more clearly.
Input
There are several test cases, please process till EOF.
For each test case, the first line is an integer N (1 <= N <= 100000), the number of enemy’s power bases. Each of the next three lines contains 3 integers, x, y, z, denoting the coordinates of S, T, and the fixed shooting direction. The last N lines contains 9 integers, x1, y1, z1, x2, y2, z2, x3, y3, z3, denoting the coordinates of the three vertices of enemy’s power base. It is guaranteed that all the triangles will not degenerate.
And the absolute value of all numbers except N and T will not exceed 1000.
Output
For each test case, print the expected number of destroyed power bases.
Any answer within an absolute error less than or equal to 10-6 would be accepted.
Sample Input
2
0 0 0
2 0 0
0 0 1
-1 0 1 1 0 1 -1 0 2
1 1 -1 1 -1 -1 2 0 -1
Sample Output
1.00000000
总结:这是一道计算几何,保持我队优良传统弃了它,何时等到可以进区域赛会重新看它的
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
#define PB push_back
#define MP make_pair
#define AA first
#define BB second
#define OP begin()
#define ED end()
#define SZ size()
#define SORT(x) sort(x.OP,x.ED)
#define SQ(x) ((x)*(x))
#define SSP system("pause")
#define cmin(x,y) x=min(x,y)
#define cmax(x,y) x=max(x,y)
typedef long long LL;
typedef pair<int, int> PII;
const double eps=1e-8;
const double PI=acos(-1.);
const LL MOD = 1000000007;
int sign(double x) {return x<-eps?-1:x>eps;}
struct spt {
double x,y,z;
spt(double _x=0,double _y=0,double _z=0) :x(_x),y(_y),z(_z) {}
spt operator + (spt s) {return spt(x+s.x,y+s.y,z+s.z);}
spt operator - (spt s) {return spt(x-s.x,y-s.y,z-s.z);}
spt operator * (double s) {return spt(x*s,y*s,z*s);}
spt operator / (double s) {return spt(x/s,y/s,z/s);}
double len() const {return sqrt(SQ(x)+SQ(y)+SQ(z));}
double operator * (spt s) {return x*s.x+y*s.y+z*s.z;} //点积dot
spt operator ^ (spt s) { //叉积det
spt ret;
ret.x=y*s.z-z*s.y;
ret.y=z*s.x-x*s.z;
ret.z=x*s.y-y*s.x;
return ret;
}
void output(char *s="") {printf("%s:%.6f %.6f %.6f\n",s,x,y,z);}
void input() {scanf("%lf%lf%lf",&x,&y,&z);}
} Orz(0,0,0);
spt S,T,V,A,B,C;
double disLP(spt p1,spt p2,spt q) {
return fabs((p2-p1)*(q-p1)/(p2-p1).len());
}
double disLL(spt p1,spt p2,spt q1,spt q2) {
spt p=q1-p1,u=p2-p1,v=q2-q1;
double d=(u*u)*(v*v)-SQ(u*v);
if(sign(d)==0)return disLP(q1,q2,p1);
double s=((p*u)*(v*v)-(p*v)*(u*v))/d;
return disLP(q1,q2,p1+u*s);
}
int isFL(spt p,spt o,spt q1,spt q2,spt &is) {
double a=o*(q2-p),b=o*(q1-p);
double d=a-b;
if(sign(d)==0)return 0;
is=(q1*a-q2*b)/d;
return 1;
}
int isFF(spt p1,spt o1,spt p2,spt o2,spt &ip,spt &io) {
spt e=o1^o2;
spt v=o1^e;
double d=o2*v;
if(sign(d)==0)return 0;
ip=p1+v*(o2*(p2-p1))/d,io=e;
return 1;
}
int inner(spt O,spt A,spt B,spt C) {
double S=((B-A)^(C-A)).len();
double S1=((A-O)^(B-O)).len();
double S2=((A-O)^(C-O)).len();
double S3=((C-O)^(B-O)).len();
return sign(S-S1-S2-S3)==0;
}
int inner(double o,double a,double b) {
return sign(max(a,b)-o)>=0&&sign(min(a,b)-o)<=0;
}
int inner(spt O,spt A,spt B) {
return inner(O.x,A.x,B.x)&&inner(O.y,A.y,B.y)&&inner(O.z,A.z,B.z);
}
int main() {
//freopen("","r",stdin);
//freopen("","w",stdout);
int i,j,k,u,v,w,p,q,r,n,m;
while(~scanf("%d",&n)) {
S.input(),T.input();
V.input();
double ans=0;
spt U= (S-T) ^V;
for(j=0; j<n; j++) {
A.input(),B.input(),C.input();
spt D= (B-A) ^ (C-A);
if(sign(D.len()) ==0) continue;
if(sign(U.len())==0) {
//TODO S->T Yu V PingXing
spt is;
int f=isFL(A,D,S,S+V,is);
if(f) {
ans+=inner(is,A,B,C);
continue;
}
if(sign((S-A)*D))continue;
spt iAB,iBC,iAC;
int fAB=isFL(A,D^(A-B),S,S+V,iAB);
int fBC=isFL(B,D^(B-C),S,S+V,iBC);
int fAC=isFL(C,D^(C-A),S,S+V,iAC);
fAB&=inner(iAB,A,B);
fBC&=inner(iBC,B,C);
fAC&=inner(iAC,A,C);
ans+=fAB|fBC|fAC;
continue;
}
if(sign(V*D)==0) {
if(sign((S-A)*D)==0&&sign((T-A)*D)==0) {
//TODO V//ABC && STABC on flat
spt iA,iB,iC;
int fA=isFL(S,(T-S)^D,A,A+V,iA);
int fB=isFL(S,(T-S)^D,B,B+V,iB);
int fC=isFL(S,(T-S)^D,C,C+V,iC);
double len=(T-S).len();
double le=0,re=len;
vector<double>L;
if(fA)L.PB((iA-S)*(T-S)/len);
if(fB)L.PB((iB-S)*(T-S)/len);
if(fC)L.PB((iC-S)*(T-S)/len);
sort(L.OP,L.ED);
if(L.SZ<2)continue;
double pe=L[0],qe=L[L.SZ-1];
cmax(pe,le);
cmin(qe,re);
if(qe>pe)ans+=(qe-pe)/len;
}
continue;
}
spt SP,TP,iAB,iBC,iAC;
assert(isFL(A,D,S,S+V,SP));
assert(isFL(A,D,T,T+V,TP));
if(inner(SP,A,B,C)&&inner(TP,A,B,C)) {ans+=1; continue;}
vector<spt>L;
L.PB(SP),L.PB(TP);
int fAB=isFL(A,D^(A-B),SP,TP,iAB);
int fBC=isFL(B,D^(B-C),SP,TP,iBC);
int fAC=isFL(C,D^(C-A),SP,TP,iAC);
double len=(SP-TP).len();
if(fAB&&inner(iAB,SP,TP))for(i=0; i+1<L.SZ; i++)
if(inner(iAB,L[i],L[i+1])) {L.insert(L.OP+i+1,iAB); break;}
if(fBC&&inner(iBC,SP,TP))for(i=0; i+1<L.SZ; i++)
if(inner(iBC,L[i],L[i+1])) {L.insert(L.OP+i+1,iBC); break;}
if(fAC&&inner(iAC,SP,TP))for(i=0; i+1<L.SZ; i++)
if(inner(iAC,L[i],L[i+1])) {L.insert(L.OP+i+1,iAC); break;}
for(i=0; i+1<L.SZ; i++) {
spt mid=(L[i]+L[i+1])/2;
if(inner(mid,A,B,C))
ans+=(L[i+1]-L[i]).len()/len;
}
}
printf("%.8f\n",ans);
}
return 0;
}
E. Circular Lamps(HDU4806)