跟UVA 674 Coin Change 一样。就是范围变大了而已。
不过当结果只有一种可能的时候需要输出 There is only 1 way to produce n cents change.
int 会溢出,使用 long long
#include<cstdio> #include<cstring> #include<string> #include<queue> #include<algorithm> #include<map> #include<stack> #include<iostream> #include<list> #include<set> #include<vector> #include<cmath> #define INF 0x7fffffff #define eps 1e-8 #define LL long long #define PI 3.141592654 #define CLR(a,b) memset(a,b,sizeof(a)) #define FOR(i,a,n) for(int i= a;i< n ;i++) #define FOR0(i,a,b) for(int i=a;i>=b;i--) #define pb push_back #define mp make_pair #define ft first #define sd second #define sf scanf #define pf printf #define acfun std::ios::sync_with_stdio(false) #define SIZE 30000 +1 using namespace std; int main() { LL dp[SIZE]; CLR(dp,0); dp[0]=1; int coin[]={1,5,10,25,50}; FOR(i,0,5) { FOR(j,coin[i],SIZE) dp[j]+=dp[j-coin[i]]; } // FOR(i,0,SIZE) // pf("%d %lld\n",i,dp[i]); int n; while(~sf("%d",&n)) { if(dp[n]==1) pf("There is only 1 way to produce %d cents change.\n",n); else pf("There are %lld ways to produce %d cents change.\n",dp[n],n); } }
原文地址:http://blog.csdn.net/dongshimou/article/details/40856363