2014
02-17

God of War

At 184~280 A.D ,there were many kingdoms in China. Three strongest among them are "Wei", "Shu", "Wu". People call this period as "Three Kingdoms".
HH is a super "Three Kingdoms" fan, because at this period there were many heroes and exciting stories. Among the heroes HH worships LvBu most.
LvBu is the God of War and is also intelligent, but his ambition is too big while enemies are too powerful .Many monarchs wanted to kill him.
At 198 A.D ,CaoCao fought with LvBu at Xuzhou.Though Lvbu is the God of War ,CaoCao had so many generals: Xuchu,DianWei XiahouChun……Facing so many heroes ,could LvBu beat all of them?

Given the LvBu’s ATI, DEF, HP, and enemies’ ATI, DEF,HP, experience (if LvBu killed one of his enemies, he can get that experience ,and if his experience got more than or equal to 100*level,he would level-up and become stronger) and the In_ATI,In_DEF,In_HP(indicating when LvBu levels up,his ability will increase this point).
Each turn LvBu will choose an enemy to fight. Please help LvBu find a way to beat all of enemies and survive with the max HP.
Here’s a fight between LvBu and A:
If LvBu attack A, A will lose Max(1,LvBu’s ATI- A’s DEF) hp;
If A survived, he will give LvBu Max(1,A’ATI- LvBu’DEF) injury.
If LvBu is still alive, repeat it untill someone is dead(hp <= 0).

LvBu’s initial level is 1 and experience is 0,and he can level up many times.

The input contains at most 20 test cases.
For each case , the first line contains six intergers ,indicating LvBu’s ATI,DEF,HP and In_ATI,In_DEF,In_HP.
The next line gives an interger N(0<N<=20),indicating the number of the enemies .
Then N lines followed, every line contains the name(the length of each name is no more than 20),ATI,DEF,HP, experience(1<experience<=100).

The input contains at most 20 test cases.
For each case , the first line contains six intergers ,indicating LvBu’s ATI,DEF,HP and In_ATI,In_DEF,In_HP.
The next line gives an interger N(0<N<=20),indicating the number of the enemies .
Then N lines followed, every line contains the name(the length of each name is no more than 20),ATI,DEF,HP, experience(1<experience<=100).

100  80  100  5  5  5
2
ZhangFei 95  75  100  100
XuChu 90  90  100  90

100 75 100 5 5 5
1
GuanYu 95 85 100 100

30
Poor LvBu,his period was gone.

100  80  100  5  5  5
2
ZhangFei 95  75  100  100
XuChu 90  90  100  90

100 75 100 5 5 5
1
GuanYu 95 85 100 100

C艹代码：

#include <stdio.h>
#include <string.h>
#define MIN 21
#define MAX 1<<21
#define max(a,b) (a)>(b)?(a):(b)

struct node {

int at,def,hp,ex;
int n,m,state,ans;

void State_Dp() {

int i,j,k,meat,heat,mtime,htime;
int curhp,curex,curat,curdef;

for (i = 0; i < (1<<n); ++i)
for (j = 0; j < n; ++j)
if (!(i & (1<<j)) && dp[i].hp) {

meat = max(1,dp[i].at - enemy[j].def);          //我们打对方多少hp
heat = max(1,enemy[j].at - dp[i].def);          //对方打我们多少hp
mtime = (dp[i].hp + heat - 1) / heat;           //我们能撑住几次攻击
htime = (enemy[j].hp + meat - 1) / meat;        //对方能撑住几次攻击

if (mtime < htime) continue;                  //我们先倒下
curhp = dp[i].hp - (htime - 1) * heat;        //攻击完的状态
curex = dp[i].ex + enemy[j].ex;
curat = dp[i].at,curdef = dp[i].def;
if (curex >= 100) {                           //升级

curex -= 100;
}

if (curhp >= dp[i|(1<<j)].hp) {             //更新答案

dp[i|(1<<j)].hp = curhp;
dp[i|(1<<j)].ex = curex;
dp[i|(1<<j)].at = curat;
dp[i|(1<<j)].def = curdef;
}
}
}

int main()
{
int i,j,k;
char tp[30];

while (scanf("%d%d%d",&dp[0].at,&dp[0].def,&dp[0].hp) != EOF) {

dp[0].ex = 0;
scanf("%d",&n);
for (i = 0; i < n; ++i)
scanf("%s %d%d%d%d",tp,&enemy[i].at,&enemy[i].def,&enemy[i].hp,&enemy[i].ex);

for (i = 1; i < (1<<n); ++i) dp[i].hp = 0;
State_Dp();
if (dp[(1<<n)-1].hp) printf("%d\n",dp[(1<<n)-1].hp);
else printf("Poor LvBu,his period was gone.\n");
}
}

1. #include <cstdio>

int main() {