HDU 1085 Holding Bin-Laden Captive!
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题目地址:点击打开链接
思路:挺不错的一题,总共写了六种代码,运行时间一次比一次少,最次的代码思路正确,只不过写废了就不贴了,最次的代码的思想的把单个硬币能表示出来的敲出来,接着再组合,而且每次循环到8001,很容易超时,母函数为(1+x^1+x^2+..x^n)(1+x^2+x^4+..+x^2n)(1+x^5+x^10+..+x^5n)其中n为每个种类硬币的数量,剩下的5种代码简介如下
AC代码1:
#include <iostream>#include<cstring>using namespace std;int c1[8005],c2[8005];int type[3] = {1,2,5};int num[3];int main(){ int i,j,k,max; while(cin>>num[0]>>num[1]>>num[2]) { max = 0; if(num[0] + num[1] + num[2] == 0) break; memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); c1[0] = 1; for(i=0; i<3; i++)//从第一个式子开始搞,并且使c1[0]=0,把能通过一种硬币表示的钱表示出来 { max += type[i] * num[i];//通过max控制不必要的循环 for(j=0; j<=max; j++) { for(k=0; k<=num[i] && j+k*type[i]<=max; k++) { c2[j+k*type[i]] += c1[j]; } } memcpy(c1,c2,sizeof(c2)); memset(c2,0,sizeof(c2)); } for(i=1;;i++) { if(c1[i] == 0) { cout<<i<<endl; break; } } } return 0;}
AC代码2:
#include <iostream>#include<cstring>using namespace std;int c1[8005],c2[8005];int type[3] = {1,2,5};int num[3];int main(){ int i,j,k,max1,max2; while(cin>>num[0]>>num[1]>>num[2]) { max2 = 0; max1 = 0; if(num[0] + num[1] + num[2] == 0) break; memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); c1[0] = 1; for(i=0; i<3; i++) { max1 = max2; max2 += type[i] * num[i]; for(j=0; j<=max1; j++)//上一个阶段的max其实还是有不必要的成分, { for(k=0; k<=num[i] && j+k*type[i]<=max2; k++) { c2[j+k*type[i]] += c1[j]; } } memcpy(c1,c2,sizeof(c2)); memset(c2,0,sizeof(c2)); } for(i=1;;i++) { if(c1[i] == 0) { cout<<i<<endl; break; } } } return 0;}
AC代码3:
#include <iostream>#include<cstring>using namespace std;int c1[8005],c2[8005];int type[3] = {1,2,5};int num[3];int main(){ int i,j,k,max1,max2; while(cin>>num[0]>>num[1]>>num[2]) { max2 = 0; max1 = 0; if(num[0] + num[1] + num[2] == 0) break; memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); c1[0] = 1; for(i=0; i<3; i++) { max1 = max2; max2 += type[i] * num[i]; for(j=0; j<=max1; j++) { if(c1[j])//c1[j]为0时相乘是没有意义的,所以加个条件,节省不少时间 { for(k=0; k<=num[i] && j+k*type[i]<=max2; k++) { c2[j+k*type[i]] += c1[j]; } } } memcpy(c1,c2,sizeof(c2)); memset(c2,0,sizeof(c2)); } for(i=1;;i++) { if(c1[i] == 0) { cout<<i<<endl; break; } } } return 0;}
这个和第三个代码差不多,改动的地方就是,只要能取到这个数就把这个数标记一下就行,不用求种类数
AC代码4:
#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>using namespace std;int c1[8001],c2[8001];int main(){ int i,j,k; int type[3] = {1,2,5}; int num[3]; while(scanf("%d%d%d",&num[0],&num[1],&num[2])) { if(num[0] + num[1] + num[2] == 0) break; memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); int max1 ,max2 = 0; c1[0] = 1;//这样赋值是把能用1块硬币表示的钱数找出来 for(i=0; i<3; i++) { max1 = max2; max2 += num[i] * type[i]; for(j=0; j<=max1; j++) { for(k=0; k<=num[i] && j + k * type[i] <= max2; k++) { if(c1[j]) c2[j+k*type[i]] = 1;//标记一下就行 } } memcpy(c1,c2,sizeof(c2)); memset(c2,0,sizeof(c2)); } for(i=1;;i++) { if(!c1[i]) { printf("%d\n",i); break; } } } return 0;}
AC代码5:参考的大神的代码:原文地址如下点击打开链接
#include<iostream>using namespace std;int main(){ int a,b,c; while(cin>>a>>b>>c) { if(a + b + c == 0) break; if(a == 0) { cout<<"1"<<endl; } else if(a + 2 * b < 4) { cout<<a+2*b+1<<endl; } else cout<<a+2*b+5*c+1<<endl; } return 0;}
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