大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。
Holding Bin-Laden Captive!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15384 Accepted Submission(s): 6892
“Oh, God! How terrible! ”
Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (硬币) (three kinds– 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!
1 1 3 0 0 0
4仍旧是母函数水过。题意：有3种面值的硬币{1,2,5} 如今给出这3种硬币的个数，求最小不能组成的面值。。
暴力生成 a[]数组扫一遍第一个为0的就是最小不能组成的面值#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cctype> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> #include <list> #define maxn 100100 #define ll long long #define INF 0x3f3f3f3f #define pp pair<int,int> using namespace std; int a[maxn],b[maxn],v[3]={1,2,5},p,n[3]; void solve() { p=n[0]+n[1]*2+n[2]*5; memset(a,0,sizeof(a)); a[0]=1; for(int i=0;i<3;i++) { memset(b,0,sizeof(b)); for(int j=0;j<=n[i]&&j*v[i]<=p;j++) for(int k=0;k+j*v[i]<=p;k++) b[k+j*v[i]]+=a[k]; memcpy(a,b,sizeof(b)); } int ans; for(ans=0;ans<=p;++ans) if(a[ans]==0)break; printf("%d\n",ans); } int main() { while(~scanf("%d%d%d",&n[0],&n[1],&n[2])) { if(!n[0]&&!n[1]&&!n[2])break; solve(); } return 0; }
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