Time Limit:2000MS Memory Limit:65535K
Type: Program Language: Not Limited
Description
Lys plays Love Live game now. You can level up via playing songs and get experiences ei but consume spirit si. Initially, you have n songs and spirit SP, empty experience. When you get enough experience, you step in next level, and the experience you got flush to empty and the spirit will be filled full. What’s more, when you step in next level, the total spirit SP will increase c, means you have c extra spirit to consume, and required q more experiences to step in next level. Now give you n songs, and the experience you can get and the spirit you should consume of each song. The initially spirit SP you have, the first level experience requirement. You can tell how the level you can step in?
Input
First line has one integer t, the number cases. For each case, first line, n(1<=n<=10), n songs, SP(1<=SP<=1000), the initial spirit, EP(1<=EP<=100000), the first level requirement experiences, c(1<=c<=100), the extra spirit you can get for each level, q(1<=q<=100), the extra requirement experiences for each level. Next n lines, for each line, has two integers, s, e, consume spirit s and get experiences e for each song.
Output
For each case, print the most level you can get. If the level is larger than 10000, you should only output 10000.
Sample Input
1 2 10 10 5 6 3 3 4 4
Sample Output
2
Hint
Before playing the song, you have 10 spirit, and require 10 experience to step into next level. You can play the first song two times and the second song one time, consume 10 spirt, and get 10 experiences, step level 2. And spirt become 15, and require 16 experiences to next level. Then you can not step into next level with this spirit.
思路:完全背包,每一次在背包容量为sp时,获得的最大价值为mv,当mv大于等于ep时,表示能升级,此时背包容量扩充为 sp + c, 升级条件变为 mv >= (ep + q)
而随着背包容量的扩充,之前的dp[]已经保存了对应状态的最优值,故不必重新dp一遍
1 /* 2 times 108ms 3 by orc 4 */ 5 #include <cstdio> 6 #include <iostream> 7 #include <cstring> 8 #include <algorithm> 9 #include <queue> 10 #include <set> 11 using namespace std ; 12 int n, sp, ep, c, q ; 13 int s[15], e[15] ; 14 int nsize ; 15 int dp[1100000] ; 16 int getans(int cur) 17 { 18 int& res = dp[cur] ; 19 if(res != -1) return res ; 20 res = 0 ; 21 for(int i = 1; i <= n; ++i) 22 if(cur >= s[i]) 23 res = max(res,getans(cur - s[i]) + e[i]) ; 24 return res ; 25 } 26 int main() 27 { 28 #ifdef LOCAL 29 freopen("in.txt","r",stdin) ; 30 #endif 31 32 int t ; 33 scanf("%d",&t) ; 34 while(t--) 35 { 36 scanf("%d%d%d%d%d",&n,&sp,&ep,&c,&q) ; 37 for(int i = 1 ; i <= n; ++i) scanf("%d%d",&s[i],&e[i]) ; 38 int nsize = sp, lev = 1; 39 memset(dp, -1 ,sizeof dp) ; 40 while(1) 41 { 42 int now = getans(nsize) ; 43 // printf("[%d] ",now) ; 44 if(now >= ep) {lev++; nsize += c ; ep += q ;} 45 46 else break ; 47 if(lev >= 10000) break ; 48 } 49 if(lev >= 10000) printf("10000 ") ; 50 else printf("%d ",lev) ; 51 } 52 53 }