Hdu3591the trouble of Xiaoqian multiple backpack + Full backpack

//Give the value of Xiaoqian coins and the number of each kind of money//Merchant's number of each kind of money is infinite, Xiaoqian pay 20000//Ask how to pay the minimum number of coins in the transaction//xiaoqian is a multi-pack .//Merchant is full backpack#include <cstdio>#include <cstring>#include <iostream>using namespace STD;Const intMAXN =20010;Const intINF =0x3f3f3f3f;intDP[MAXN];intDP_1[MAXN];intC[MAXN],V[MAXN];intN, T;intCAS =0;intMain () {//freopen ("In.txt", "R", stdin);     while(scanf("%d%d", &n, &t) && (n+t)) { for(inti =1; I <= n;i++)scanf("%d", &v[i]); for(inti =1; I <= n;i++)scanf("%d", &c[i]); for(inti =1; i < maxn;i++) dp[i] = dp_1[i] = inf; dp_1[0] =0; dp[0] =0; for(inti =1; I <= n;i++) for(intj = V[i]; J < Maxn; J + +) Dp_1[j] = min (Dp_1[j], Dp_1[j-v[i]] +1) ; for(inti =1; I <= n;i++) { for(intK =1; k <= c[i];k*=2)            { for(intj = MAXN-1; J >= k*v[i];j--) dp[j] = min (Dp[j], Dp[j-k*v[i]] + k);            C[i]-=k; } for(intj = maxn-1; J >= c[i]*v[i];j--) dp[j] = min (Dp[j], Dp[j-c[i]*v[i]] + c[i]); }intans = inf; for(inti = t; I <= MAXN-Ten; i++)if(Dp[i]! = INF && dp_1[i-t]! = inf) ans = min (dp[i] + dp[i-t], ans);printf("Case%d:", ++cas);if(ans = = inf)puts("-1") ;Else cout<<ans<<endl; }}

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Hdu3591the trouble of Xiaoqian multiple backpack + Full backpack