HDU_1085 Holding Bin-Laden Captive!(数学)
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题目请点我
思路:
一开始用动态规划做的,但是其实只是一道证明题。
公式:
1.若0~4能取到,则不能取到的数为num = 5*s5 + 2*s2+s1;
2.num = 0~4不能取到的数;
代码:
dp版本:
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#define MAX 8010using namespace std;int a[3]={1,2,5};int num[3];int dp[MAX];int main(){ int p,q,w,sum; while( scanf("%d%d%d",&p,&q,&w)!=EOF ) { if( p == 0 && q == 0 && w == 0 ) break; sum = 0; num[2] = w; num[1] = q; num[0] = p; memset(dp,0,sizeof(dp)); dp[0] = 1; for(int i = 0; i < 3; i++) sum += num[i]*a[i]; for( int i = 0; i < 3; i++ ){ for( int j = sum; j >= 0; j-- ){ if( dp[j] == 1 ){ for( int t = 1; t <= num[i]; t++ ){ dp[j+t*a[i]] = 1; } } } } for( int i = 0; i <= sum+1; i++ ){ if( dp[i] == 0 ){ printf("%d\n",i); break; } } } return 0;}
数学版本:
#include <iostream>#include <cstdio>#include <cstdlib>using namespace std;int main(){ int n1,n2,n3; while( scanf("%d%d%d",&n1,&n2,&n3)!=EOF ){ if( n1 == 0 && n2 == 0 && n3 == 0 ){ break; } for( int i = 1; i <= 5; i++ ){ if( i == 1 ){ if( n1 == 0 ){ printf("%d\n",i); break; } } else if( i < 5 && i >= 2 ){ if( n2*2 + n1 < i ){ printf("%d\n",i); break; } } else{ printf("%d\n",5*n3+2*n2+1*n1+1); break; } } }}
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