Square Coins

Description


People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland. 
There are four combinations of coins to pay ten credits: 


ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin. 


Your mission is to count the number of ways to pay a given amount using coins of Silverland.
Input
The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.
Output
For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output.
Sample Input
2
10
30
0
Sample Output
1
4
27
Source
Asia 1999
Uploader

crq

#include<stdio.h>

#define max 310
int main()
{
    int coin[17] = {1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289}; //列出所有coin的面值
    int n,i,j,k, an[max], bn[max]; //an[]记录结果,bn[]记录中间值
    while(scanf("%d", &n) && n)
    {
        for(i = 0; i <= n; i ++) //赋初值
        {
            an[i] = 1;
            bn[i] = 0;
        }
        for(j = 1; j < 17; j ++) 
        {
            for(i = 0; i <= n; i ++)
            {
                for(k = 0; k + i <= n; k += coin[j])
                    bn[k + i] += an[i];
            }
            for(i = 0; i <= n; i ++)
            {
                an[i] = bn[i];
                bn[i] = 0;
            }
        }
        printf("%d\n", an[n]);
    }
    return 0;
}



clc; clear all; [filename,pathname]=uigetfile({'*.jpg;*.tif;*.png;*gif','all imagine files';'*.*','all files'},'select your photo'); path=[pathname,filename]; image=imread(path); % axes(handles.photo); imshow(image);%显示图片 %image processing I=rgb2gray(image); I=rangefilt(I);%滤波 background = imopen(I,strel('disk',11));%使用形态学开来估计背景 I2 = I-background;%从原始图像中减去背景图像 I3 = imadjust(I2);%增强对比度 bw = imbinarize(I3);%阈 值图像 bw = bwareaopen(bw,160);%降噪150,160 bw=edge(bw,'canny'); %边缘检测 %bw=1-bw; % axes(handles.a1); imshow(bw); %se=strel('disk',13);%15 se=strel('square',15);%15 bw1=imclose(bw,se);%闭 bw2=imdilate(bw1,se);%膨胀 bw2=imerode(bw2,se);%腐蚀 bw3=imfill(bw2,'holes'); % axes(handles.a2); imshow(bw3); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %circle detection rmin = 20; rmax = 2500; radiusRange=[rmin rmax]; [center, rad] = imfindcircles(bw3,radiusRange,'EdgeThreshold',0.13);%检测灵敏度(边缘渐变阈值)0.3 display(center); display(rad); % axes(handles.a3); imshow(bw3); viscircles(center, rad,'Color','b'); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %initialize the number of coins one=0; half=0; little=0; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %coin recognition [m,n]=size(rad); num=m; i=1; j=num; min=rad(i); max=rad(j); while i<=j if rad(i)<rad(j) if rad(i)<min min=rad(i); else if rad(j)<max max=rad(j); end end else if rad(j)<min min=rad(j); else if rad(i)<max max=rad(i); end end end i=i+1; j=j-1; end sum=0; for i=1:num sum=rad(i)+sum; end % ave=(sum-(min+max))/(num-2); ave = sum/num; for i=1:num if 0.6<(rad(i)/ave)&&(rad(i)/ave)<1.5 if rad(i)>ave one=one+1; else if 0.93<(rad(i)/ave) && rad(i)<=ave half=half+1; else little = little+1; end end end end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %display results sum=half*0.5+one+little*0.1; one half little sum这段代码什么意思
06-12
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