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NENU Summer Training 2014 #8
Ignatius
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user‘s result file, then the system compare the two files. If the two files are absolutly same, then the Judge System return "Accepted", else if the only differences between the two files are spaces(‘ ‘), tabs(‘\t‘), or enters(‘\n‘), the Judge System should return "Presentation Error", else the system will return "Wrong Answer".
Given the data of correct output file and the data of user‘s result file, your task is to determine which result the Judge System will return.
Given the data of correct output file and the data of user‘s result file, your task is to determine which result the Judge System will return.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case has two parts, the data of correct output file and the data of the user‘s result file. Both of them are starts with a single line contains a string "START" and end with a single line contains a string "END", these two strings are not the data. In other words, the data is between the two strings. The data will at most 5000 characters.
Each test case has two parts, the data of correct output file and the data of the user‘s result file. Both of them are starts with a single line contains a string "START" and end with a single line contains a string "END", these two strings are not the data. In other words, the data is between the two strings. The data will at most 5000 characters.
Output
For each test cases, you should output the the result Judge System should return.
Sample Input
4 START 1 + 2 = 3 END START 1+2=3 END START 1 + 2 = 3 END START 1 + 2 = 3 END START 1 + 2 = 3 END START 1 + 2 = 4 END START 1 + 2 = 3 END START 1 + 2 = 3 END
Sample Output
Presentation Error Presentation Error Wrong Answer Presentation Error
分析：
最好的方法就是将输入的数组去除空行 \n,\t然后再两两进行比较
代码：
#include<stdio.h>#include<string.h>const int maxn=5001;char s1[maxn],s2[maxn];void input(char *s){ char temp[maxn]; gets(temp); while(strcmp(temp,"START")!=0) gets(temp); while(gets(temp)) { if(strcmp(temp,"END")==0) break; if(strlen(temp)!=0) strcat(s,temp); strcat(s,"\n"); }}void del(char *s,int l){ char temp[maxn]; int i,j; j=0; for(i=0;i<l;i++) { if(s[i]!=‘ ‘&&s[i]!=‘\n‘&&s[i]!=‘\t‘) temp[j++]=s[i]; } temp[j]=‘\0‘; strcpy(s,temp);}int sum(){ int l1,l2; l1=strlen(s1); l2=strlen(s2); if(l1==l2&&strcmp(s1,s2)==0) return 1; del(s1,l1); del(s2,l2); if(strcmp(s1,s2)==0) return 0; else return -1;}int main(){ int T,count; scanf("%d",&T); while(T--) { input(s1); input(s2); count=sum(); if(count==1) printf("Accepted\n"); else if(count==0) printf("Presentation Error\n"); else printf("Wrong Answer\n"); } return 0;}
#include<stdio.h>#include<string.h>const int maxn=5001;char s1[maxn],s2[maxn],t1[maxn],t2[maxn];char temp[maxn];void input(char *s,char *t){ gets(temp); while(strcmp(temp,"START")!=0) gets(temp); while(gets(temp)) { if(strcmp(temp,"END")==0) break; if(strlen(temp)!=0) strcat(s,temp); strcat(s,"\n"); } int k=0; int l=strlen(s); int i; for(i=0;i<l;i++) { if(s[i]!=‘ ‘&&s[i]!=‘\n‘&&s[i]!=‘\t‘) t[k++]=s[i]; } t[k]=‘\0‘;}int main(){ int T; scanf("%d",&T); while(T--) { s1[0]=‘\0‘; s2[0]=‘\0‘; input(s1,t1); input(s2,t2); if(strcmp(s1,s2)==0) printf("Accepted\n"); else if(strcmp(t1,t2)==0) printf("Presentation Error\n"); else printf("Wrong Answer\n"); } return 0;}
#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <vector>#include <cmath>#include <cstdio>using namespace std;#ifdef __int64typedef __int64 LL;#elsetypedef long long LL;#endifconst int inf=0x3f3f3f3f;const int maxn=5050;char t[maxn],as[maxn],bs[maxn],ap[maxn],bp[maxn];int main(){ int n,ca,cb; scanf("%d",&n); getchar(); while(n--) { gets(t); memset(t,0,sizeof(t)); memset(as,0,sizeof(as)); memset(bs,0,sizeof(bs)); memset(ap,0,sizeof(ap)); memset(bp,0,sizeof(bp)); ca=cb=0; while(gets(t)) { if(strcmp(t,"END")==0) break; strcat(as,t); ca++; } gets(t); while(gets(t)) { if(strcmp(t,"END")==0) break; strcat(bs,t); cb++; } if(ca==cb&&strcmp(as,bs)==0) printf("Accepted\n"); else { int la=strlen(as),lb=strlen(bs); int lla=0,llb=0; for(int i=0;i<la;i++) { if(as[i]!=‘ ‘&&as[i]!=‘\n‘&&as[i]!=‘\t‘) ap[lla++]=as[i]; } for(int i=0;i<lb;i++) { if(bs[i]!=‘ ‘&&bs[i]!=‘\n‘&&bs[i]!=‘\t‘) bp[llb++]=bs[i]; } if(strcmp(ap,bp)==0) printf("Presentation Error\n"); else printf("Wrong Answer\n"); } } return 0;}
Parabola
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
Ignatius bought a land last week, but he didn‘t know the area of the land because the land is enclosed by a parabola and a straight line. The picture below shows the area. Now given all the intersectant points shows in the picture, can you tell Ignatius the area of the land?
Note: The point P1 in the picture is the vertex of the parabola.
Note: The point P1 in the picture is the vertex of the parabola.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains three intersectant points which shows in the picture, they are given in the order of P1, P2, P3. Each point is described by two floating-point numbers X and Y(0.0<=X,Y<=1000.0).
Each test case contains three intersectant points which shows in the picture, they are given in the order of P1, P2, P3. Each point is described by two floating-point numbers X and Y(0.0<=X,Y<=1000.0).
Output
For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.
Sample Input
2 5.000000 5.000000 0.000000 0.000000 10.000000 0.000000 10.000000 10.000000 1.000000 1.000000 14.000000 8.222222
Sample Output
33.33 40.69
Hint
For float may be not accurate enough, please use double instead of float.
分析：
一道高数题 就是求出曲线和直线的方程在积分就好了！（曲线的方程可用顶点式 y = a(x-h)^2+l, (h,L) 为顶点坐标）
代码：
1.
设二次函数为y=a*x^2+b*x+c
（1）由于p1为二次函数顶点，所以b=-2*a*x1 （2）代入二次函数式子 1 里得y=a*x^2-2*a*x1*x+c（3）把p1,p2坐标代入 3 中，算出a，
然后由 2 算出b ,然后a、b、p3坐标代入 1 中算出c。这样下来就算出a、b、c了，那么ans=微积分求面积-梯形面积。
#include<stdio.h>int main(){ int n; double x1,x2,x3,y1,y2,y3; double s,a,b,c,k,h; while(~scanf("%d",&n)) { while(n--) { scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3); a=(y2-y1)/((x2-x1)*(x2-x1)); b=-2*a*x1; c=y1-a*x1*x1-b*x1; k=(y2-y3)/(x2-x3); h=y2-k*x2; s=(a*(x3*x3*x3-x2*x2*x2)/3)+((b-k)*(x3*x3-x2*x2)/2)+((c-h)*(x3-x2)); printf("%.2lf\n",s); } } return 0;}
2.
这里我们设抛物线顶点式方程y=a*(x-h)^2+k；顶点坐标为（h,k)。h=x1；k=y1；a=(y-k)/(x-h)^2；
由积分可得（x2，x3）之间的面积，再减去梯形的面积即可
#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <vector>#include <cstdio>#include <cmath>using namespace std; #ifdef __int64typedef __int64 LL;#elsetypedef long long LL;#endif const int inf = 0x3f3f3f3f;const int maxn = 100000; int main() { int n; double x1,y1,x2,y2,x3,y3; scanf("%d",&n); while(n--) { scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3); double b=x1; double c=y1; double a=(y2-c)/(x2-b)/(x2-b); double k=(y3-y2)/(x3-x2); double b1=y2-k*x2; double s1=a*x3*x3*x3/3-(2*a*b+k)*x3*x3/2+(a*b*b+c-b1)*x3; double s2=a*x2*x2*x2/3-(2*a*b+k)*x2*x2/2+(a*b*b+c-b1)*x2; printf("%.2f\n",s1-s2); } return 0;}
NENU Summer Training 2014 #8
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