http://www.cnblogs.com/AOQNRMGYXLMV/p/4934747.html
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 5 using namespace std; 6 const int maxn=12; 7 const int inf=0x3f3f3f3f; 8 int a[maxn]; 9 int b[maxn]; 10 int val[]={1,5,10,20,50,100,200,500,1000,2000}; 11 int Solve(int t) 12 { 13 int ans=0; 14 for(int i=9;i>=0;i--) 15 { 16 if(i==4||i==7) 17 { 18 int cnt=min(t/(val[i]*2),b[i]/2); 19 ans+=cnt*2; 20 t-=val[i]*2*cnt; 21 } 22 else 23 { 24 int cnt=min(t/val[i],b[i]); 25 ans+=cnt; 26 t-=val[i]*cnt; 27 } 28 if(t==0) 29 { 30 break; 31 } 32 } 33 if(t==0) 34 { 35 return ans; 36 } 37 else 38 { 39 return inf; 40 } 41 } 42 int main() 43 { 44 int T; 45 scanf("%d",&T); 46 while(T--) 47 { 48 int p; 49 scanf("%d",&p); 50 int tot=0; 51 int sum=0; 52 for(int i=0;i<10;i++) 53 { 54 scanf("%d",&a[i]); 55 tot+=a[i]; 56 sum+=val[i]*a[i]; 57 } 58 if(sum<p) 59 { 60 printf("-1 "); 61 continue; 62 } 63 p=sum-p; 64 int ans=inf; 65 for(int i=0;i<2;i++) 66 { 67 for(int k=0;k<2;k++) 68 { 69 int temp=p; 70 memcpy(b,a,sizeof(a)); 71 if(i) 72 { 73 if(b[4]) 74 { 75 temp-=val[4]; 76 b[4]--; 77 } 78 else 79 { 80 continue; 81 } 82 } 83 if(k) 84 { 85 if(b[7]) 86 { 87 temp-=val[7]; 88 b[7]--; 89 } 90 else 91 { 92 continue; 93 } 94 } 95 if(temp>=0) 96 { 97 ans=min(ans,Solve(temp)+i+k); 98 } 99 } 100 } 101 if(ans==inf) 102 { 103 printf("-1 "); 104 } 105 else 106 { 107 printf("%d ",tot-ans); 108 } 109 } 110 return 0; 111 }
考虑问题的反面，求最大转化为求剩余的最小。然后从大到小贪心，能多用大面值的尽量多用。特判20,50；200,500