uva 165 Stamps
2013-11-14 16:42
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题目地址:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=108&page=show_problem&problem=101
题目描述:
The government of Nova Mareterrania requires that various legal documents have stamps attached to them so that the government can derive revenue from them. In terms of recent legislation, each class of document
is limited in the number of stamps that may be attached to it. The government wishes to know how many different stamps, and of what values, they need to print to allow the widest choice of values to be made up under these conditions. Stamps are always valued
in units of $1.
This has been analysed by government mathematicians who have derived a formula for n(h,k), where h is the number of stamps that may be attached to a document, k is the
number of denominations of stamps available, and n is the largest attainable value in a continuous sequence starting from $1. For instance, if h=3, k=2 and the denominations are $1 and $4, we can make all the values from $1 to $6
(as well as $8, $9 and $12). However with the same values of h and k, but using $1 and $3 stamps we can make all the values from $1 to $7 (as well as $9). This is maximal, so n(3,2) = 7.
Unfortunately the formula relating n(h,k) to h, k and the values of the stamps has been lost--it was published in one of the government reports but no-one can remember
which one, and of the three researchers who started to search for the formula, two died of boredom and the third took a job as a lighthouse keeper because it provided more social stimulation.
The task has now been passed on to you. You doubt the existence of a formula in the first place so you decide to write a program that, for given values of h and k, will determine an optimum set
of stamps and the value of n(h,k).
to 9. (The President lost his little finger in a shooting accident and cannot count past 9).
(
题意:组合邮票 组合钞票
题解:搜索,但要剪枝搜索,不然容易TLE。
面值为1的必选(初始化),stamp[1]=1; max[1]=h;
则第x张面值得取值范围为 stamp[x-1]+1<=stamp[X]<=max[X-1]+1;
超过之前的max[X-1]+1,则面值为max[x-1+1],一定无法组合出来;
不过也可以打表来做,就那么几种情况。。。。
代码:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=108&page=show_problem&problem=101
题目描述:
Stamps |
is limited in the number of stamps that may be attached to it. The government wishes to know how many different stamps, and of what values, they need to print to allow the widest choice of values to be made up under these conditions. Stamps are always valued
in units of $1.
This has been analysed by government mathematicians who have derived a formula for n(h,k), where h is the number of stamps that may be attached to a document, k is the
number of denominations of stamps available, and n is the largest attainable value in a continuous sequence starting from $1. For instance, if h=3, k=2 and the denominations are $1 and $4, we can make all the values from $1 to $6
(as well as $8, $9 and $12). However with the same values of h and k, but using $1 and $3 stamps we can make all the values from $1 to $7 (as well as $9). This is maximal, so n(3,2) = 7.
Unfortunately the formula relating n(h,k) to h, k and the values of the stamps has been lost--it was published in one of the government reports but no-one can remember
which one, and of the three researchers who started to search for the formula, two died of boredom and the third took a job as a lighthouse keeper because it provided more social stimulation.
The task has now been passed on to you. You doubt the existence of a formula in the first place so you decide to write a program that, for given values of h and k, will determine an optimum set
of stamps and the value of n(h,k).
Input
Input will consist of several lines, each containing a value for h and k. The file will be terminated by two zeroes (0 0). For technical reasons the sum of h and k is limitedto 9. (The President lost his little finger in a shooting accident and cannot count past 9).
Output
Output will consist of a line for each value of h and k consisting of the k stamp values in ascending order right justified in fields 3 characters wide, followed by a space and an arrow(
->) and the value of n(h,k) right justified in a field 3 characters wide.
Sample input
3 2 0 0
Sample output
1 3 -> 7
题意:组合邮票 组合钞票
题解:搜索,但要剪枝搜索,不然容易TLE。
面值为1的必选(初始化),stamp[1]=1; max[1]=h;
则第x张面值得取值范围为 stamp[x-1]+1<=stamp[X]<=max[X-1]+1;
超过之前的max[X-1]+1,则面值为max[x-1+1],一定无法组合出来;
不过也可以打表来做,就那么几种情况。。。。
代码:
/* uva Stamps 165 we can print all the number to a table them use the table it's so nice that I use the print table strategy to solve the problem stamp[x-1]+1<=stamp[X]<=max[X-1]+1 max[X-1]+1 before x-1 kind of stamps must not make this value besides this value is min value they can not make if a var < value they must can make it!! */ #include<stdio.h> #include<stdlib.h> #include<string.h> #include<algorithm> #define DMAXS 32 using namespace std; int h=0,k=0; int t[8+5]={0};//t[0]*x[0]+t[1]*x[1]+.....=sum int x[8+5]={0};//the denominations values of the stamps int Sum=0;//n(h,k) int MaxSum=0;//max n(h,k) int MaxX[8+5]={0};//output for the ascending order int flagT=0; int DMAX=20; /*dfs the t array note the h number to pruning*/ int DFST(int cur) { //static int flag=0;//global var if(flagT)//pruning { return(flagT); } if(cur<=k-1) { int i=0; for(i=0;i<=h;i++) { h-=i; t[cur]=i; DFST(cur+1); h+=i; t[cur]=0; if(flagT)//pruning { return(flagT); } } } else { int Adder=0; int i=0; for(i=0;i<=k-1;i++) { Adder+=t[i]*x[i]; } if(Adder==Sum) { flagT=1; } } return(flagT); } /*calculate the t1*x1+t2*x2...==sum*/ int Calc(int Sum,int x[]) { int flag=1; flagT=0; flag=DFST(0);//dfs the t[0] return(flag); } /*DFS the x[i]'s value*/ int DFS(int cur) { if(cur<=k-1) { int i=0; for(i=x[cur-1];i<=DMAXS;i++) { x[cur]=i; DFS(cur+1); } } else//per series dfs complete { Sum=0; int flag=1; do { Sum++; flag=Calc(Sum,x);//enum the t1*x1+t2*x2...==sum }while(flag); Sum--; if(Sum>MaxSum) { MaxSum=Sum; memcpy(MaxX,x,sizeof(x)); } } return(0); } /*for test*/ int test() { for(DMAX=20;DMAX<=50;DMAX++)//max: 1 4 12 21 71 { h=5; k=4; x[0]=1; MaxSum=0; DFS(1); sort(MaxX,MaxX+k); printf("%d %d %d %d %d\n",MaxX[0],MaxX[1],MaxX[2],MaxX[3],MaxSum); } // test the [5,4] the x[i]'s max value maybe what return(0); } /*for test*/ int test1() { for(DMAX=20;DMAX<=50;DMAX++) { // h=4; // k=5;//1 3 11 15 32 70 // h=3; // k=6;//1 3 7 9 19 24 52 // h=6;//1 7 12 52 // k=3; // h=2;//1 2 5 8 11 12 13 26 // k=7; // h=1;//1 2 3 4 5 6 7 8 8 // k=8; x[0]=1; MaxSum=0; DFS(1); sort(MaxX,MaxX+k); //printf("%d %d %d %d %d\n",MaxX[0],MaxX[1],MaxX[2],MaxX[3],MaxX[4],MaxSum); int i=0; for(i=0;i<=k-1;i++) { printf("%d ",MaxX[i]); } printf("%d\n",MaxSum); } // test the [5,4] the x[i]'s max value maybe what return(0); } /* table data: 1 3 -> 7 1 2 -> 4 1 -> 3 1 3 11 18 -> 44 1 2 3 4 5 6 7 8 -> 8 1 2 5 8 11 12 13 -> 26 1 3 7 9 19 24 -> 52 1 3 11 15 32 -> 70 1 2 5 8 9 10 -> 20 1 2 3 4 5 6 -> 6 1 2 3 4 5 6 7 -> 7 1 4 6 14 15 -> 36 1 3 5 7 8 -> 16 1 2 3 4 5 -> 5 1 2 3 4 -> 4 1 3 5 6 -> 12 1 4 7 8 -> 24 */ /*main process*/ int MainProc() { while(scanf("%d%d",&h,&k)!=EOF&&h+k>0) { //table if(k==8&&h==1) { printf(" 1 2 3 4 5 6 7 8 -> 8\n"); continue; } if(k==7&&h==2) { printf(" 1 2 5 8 11 12 13 -> 26\n"); continue; } if(k==7&&h==1) { printf(" 1 2 3 4 5 6 7 -> 7\n"); continue; } if(k==6&&h==3) { printf(" 1 3 7 9 19 24 -> 52\n"); continue; } if(k==6&&h==2) { printf(" 1 2 5 8 9 10 -> 20\n"); continue; } if(k==6&&h==1) { printf(" 1 2 3 4 5 6 -> 6\n"); continue; } if(k==5&&h==4) { printf(" 1 3 11 15 32 -> 70\n"); continue; } Sum=0; x[0]=1;//the length is k,t[] is too. MaxSum=0; DFS(1); sort(MaxX,MaxX+k); int i=0; for(i=0;i<=k-1;i++) { if(MaxX[i]>=10) { printf(" "); } else { printf(" "); } printf("%d",MaxX[i]); } printf(" ->"); if(MaxSum>=10&&MaxSum<=99) { printf(" "); } else if(MaxSum<10) { printf(" "); } printf("%d\n",MaxSum); } return(0); } int main() { MainProc(); return(0); }
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