Codeforces Round #114 (Div. 1) B. Wizards and Huge Prize CF167B

考虑把得到的奖和获取的背包容量两个状态分开来算概率

结果再把符合要求的两个状态的概率相乘

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>#include <bitset>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.10f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define pll pair<ll,ll>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const ll mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  200+10;const int maxm = 40000+10;//Pretests passedint in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}struct node{    int a,b,c;    bool operator <(const node &o)const    {        if(a!=o.a)return a<o.a;        if(b!=o.b)return b<o.b;        return c<o.c;    }    bool operator == (const node &o)const    {        return a==o.a&&b==o.b&&c==o.c;    }};int pro[maxn];int pr[maxn];bool dp1[maxn][maxn];//i j  the ith - win jbool dp2[maxn][maxn][maxn];// i j k the ith - win j - had kint prizecnt;int bagcnt;double p1[maxn][maxn];//prizedouble p2[maxn][maxn][maxn];//bagint main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL    int n,l,k;    sddd(n,l,k);    r1(i,n)sd(pro[i]);    r1(i,n)sd(pr[i]);    dp1[0][0]=1;    p1[0][0]=1;    dp2[0][0][k] = 1;    p2[0][0][k] = 1;    for(int i=1; i<=n; ++i)    {        if(pr[i]<0)        {            for(int j = 0; j<=prizecnt; ++j)            {                if(dp1[prizecnt][j])                {                    if(pro[i]==0)dp1[prizecnt+1][j] = 1, p1[prizecnt+1][j] += p1[prizecnt][j];                    else if(pro[i]==100)dp1[prizecnt+1][j+1] = 1 , p1[prizecnt+1][j+1] += p1[prizecnt][j];                    else                    {                        dp1[prizecnt+1][j+1] = dp1[prizecnt+1][j] = 1;                        p1[prizecnt+1][j] += p1[prizecnt][j]*(0.01*(100-pro[i]));                        p1[prizecnt+1][j+1] += p1[prizecnt][j]*(0.01*pro[i]);                    }                }            }            prizecnt++;        }        else        {            for(int j = 0; j<=200; ++j)            {                for(int win = 0 ; win <= bagcnt; ++win)                {                    if(dp2[bagcnt][win][j])                    {                        int mx = min(200,j+pr[i]);                        if(pro[i]==0)dp2[bagcnt+1][win][j] = 1, p2[bagcnt+1][win][j] += p2[bagcnt][win][j];                        else if(pro[i]==100)dp2[bagcnt+1][win+1][mx] = 1 , p2[bagcnt+1][win+1][mx] += p2[bagcnt][win][j];                        else                        {                            dp2[bagcnt+1][win+1][mx] = dp2[bagcnt+1][win][j] = 1;                            p2[bagcnt+1][win][j] += p2[bagcnt][win][j]*(0.01*(100-pro[i]));                            p2[bagcnt+1][win+1][mx] += p2[bagcnt][win][j]*(0.01*pro[i]);                        }                    }                }            }            bagcnt++;        }    }    double ans = 0;    for(int i=0; i<=prizecnt; ++i)    {        for(int j = 0; j<=bagcnt;++j)        {            if(i+j<l)continue;            for(int bag = i; bag<=200; ++bag)                    ans+= p1[prizecnt][i]*p2[bagcnt][j][bag];        }    }    printf("%.10f",ans);    return 0;}