给出N个人的平局分X
根据GPA规则计算可能的最高平均GPA和最低平均GPA
可以DP预处理出来所有结果 或者记忆化搜索
DP:
#include "stdio.h" #include "string.h" int inf=100000000; double a[11][1100],b[11][1100]; double Max(double a,double b) { if (a<b) return b; else return a; } double Min(double a,double b) { if (a<b) return a;else return b; } int main() { int Case,i,j,x,k; for (i=0;i<=1000;i++) for (j=0;j<=10;j++) { a[j][i]=inf; b[j][i]=-1; } a[0][0]=b[0][0]=0; for (i=1;i<=10;i++) for (j=60*i;j<=100*i;j++) { for (k=60;k<=69;k++) if (j-k>=0 && a[i-1][j-k]!=inf) a[i][j]=Min(a[i][j],a[i-1][j-k]+2); for (k=70;k<=74;k++) if (j-k>=0 && a[i-1][j-k]!=inf) a[i][j]=Min(a[i][j],a[i-1][j-k]+2.5); for (k=75;k<=79;k++) if (j-k>=0 && a[i-1][j-k]!=inf) a[i][j]=Min(a[i][j],a[i-1][j-k]+3); for (k=80;k<=84;k++) if (j-k>=0 && a[i-1][j-k]!=inf) a[i][j]=Min(a[i][j],a[i-1][j-k]+3.5); for (k=85;k<=100;k++) if (j-k>=0 && a[i-1][j-k]!=inf) a[i][j]=Min(a[i][j],a[i-1][j-k]+4); } for (i=1;i<=10;i++) for (j=60*i;j<=100*i;j++) { for (k=60;k<=69;k++) if (j-k>=0 && b[i-1][j-k]!=-1) b[i][j]=Max(b[i][j],b[i-1][j-k]+2); for (k=70;k<=74;k++) if (j-k>=0 && b[i-1][j-k]!=-1) b[i][j]=Max(b[i][j],b[i-1][j-k]+2.5); for (k=75;k<=79;k++) if (j-k>=0 && b[i-1][j-k]!=-1) b[i][j]=Max(b[i][j],b[i-1][j-k]+3); for (k=80;k<=84;k++) if (j-k>=0 && b[i-1][j-k]!=-1) b[i][j]=Max(b[i][j],b[i-1][j-k]+3.5); for (k=85;k<=100;k++) if (j-k>=0 && b[i-1][j-k]!=-1) b[i][j]=Max(b[i][j],b[i-1][j-k]+4); } scanf("%d",&Case); while (Case--) { scanf("%d%d",&x,&i); printf("%.4lf %.4lf\n",a[i][i*x]/i,b[i][i*x]/i); } return 0; }
#include "stdio.h" #include "string.h" int dira[]={100,84,79,74,69}; int dirb[]={85,80,75,70,60}; double mark[]={4,3.5,3,2.5,2}; double a[1010][11],b[1010][11]; double MMin(double a,double b) { if (a<b) return a; else return b; } double MMax(double a,double b) { if (a<b) return b; else return a; } double dfs_min(int sum,int n) { double ans,temp; int i; if (a[sum][n]!=-1) return a[sum][n]; ans=500000; for (i=0;i<5;i++) { temp=sum-dira[i]; if (n!=1 && (temp/(n-1)<60 || temp/(n-1)>100)) continue; ans=MMin(ans,dfs_min(temp,n-1)+mark[i]); } return a[sum][n]=ans; } double dfs_max(int sum,int n) { double ans,temp; int i; if (b[sum][n]!=-1) return b[sum][n]; ans=0; for (i=0;i<5;i++) { temp=sum-dirb[i]; if (n!=1 && (temp/(n-1)<60 || temp/(n-1)>100)) continue; ans=MMax(ans,dfs_max(temp,n-1)+mark[i]); } return b[sum][n]=ans; } int main() { int Case,i,n,j,x; double Max,Min; scanf("%d",&Case); while (Case--) { scanf("%d%d",&x,&n); for (i=0;i<=1000;i++) for (j=0;j<=10;j++) a[i][j]=b[i][j]=-1; for (i=60;i<=69;i++) a[i][1]=b[i][1]=2; for (i=70;i<=74;i++) a[i][1]=b[i][1]=2.5; for (i=75;i<=79;i++) a[i][1]=b[i][1]=3; for (i=80;i<=84;i++) a[i][1]=b[i][1]=3.5; for (i=85;i<=100;i++) a[i][1]=b[i][1]=4; Min=Max=a[x][1]*n; Min=MMin(dfs_min(x*n,n),Min); Max=MMax(dfs_max(x*n,n),Max); printf("%.4lf %.4lf\n",Min/n,Max/n); } return 0; }
POJ 4968 DP||记忆化搜索,布布扣,bubuko.com
原文地址:http://blog.csdn.net/u011932355/article/details/38704037