标签：\n reg rap ios alt data- vector mono space
给出几道期望的例题
hdu 4405
由于种种原因,期望是要逆推的,再知道终点在哪的时候.
/*header*/ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <cmath> #define rep(i , x, p) for(int i = x;i <= p;++ i) #define sep(i , x, p) for(int i = x;i >= p;-- i) #define gc getchar() #define pc putchar #define ll long long #define mk make_pair #define fi first #define se second using std::min; using std::max; using std::swap; const int maxN = 100000 + 7; inline int gi() { int x = 0,f = 1;char c = gc; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘)f = -1;c = gc;} while(c >= ‘0‘ && c <= ‘9‘) {x = x * 10 + c - ‘0‘;c = gc;}return x * f; } void print(int x) { if(x < 0) pc(‘-‘) , x = -x; if(x >= 10) print(x / 10); pc(x % 10 + ‘0‘); } struct Node { int v , nex; }Map[maxN]; int head[maxN] , num; int tot[maxN] , to[maxN]; double f[maxN]; int n , m; double dfs(int now) { if(f[now]) return f[now]; if(now >= n) return 0.0; if(to[now]) return f[now] = dfs(to[now]); double p = 1.0 / 6; rep(i , 1, 6) f[now] += p * dfs(now + i); f[now] += 1.0; return f[now]; } void init() { memset(to , 0,sizeof(to)); rep(i , 0, n + 6) f[i] = 0.0; } int main() { while(true) { n = gi();m = gi(); if(n == 0 && m == 0) break; init(); for(int i = 1;i <= m;++ i) { int u = gi() , v = gi(); to[u] = v; } dfs(0); printf("%.4lf\n", f[0]); } return 0; }
luogu 1083
f[i][j][0/1]表示前i个教室,已经申请了j个教室,0表示这个时间段要去ci,1表示这个时间段要去di的最小的期望值
#include <iostream> #include <cstring> #include <cstdio> using namespace std; const int maxT = 2000 + 7; const int maxM = 90000 + 7; const int maxN = 300 + 7; int c[maxT],d[maxT],dis[maxN][maxN]; double k[maxT]; double E[maxT][maxT][2]; inline int read() { int x = 0,E = 1;char c = getchar(); while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘)E = -1;c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) {x = x * 10 + c - ‘0‘;c = getchar();} return x * E; } int main() { int n = read() ,m = read(),v = read(),e = read(); for(int i = 1;i <= n;++ i) c[i] = read(); for(int i = 1;i <= n;++ i) d[i] = read(); for(int i = 1;i <= n;++ i) cin >> k[i]; int u,g,w; for(int i = 1;i <= v;++ i) for(int j = 1;j < i;++ j) dis[i][j] = dis[j][i] = 999999999; for(int i = 1;i <= e;++ i) { u = read(),g = read(),w = read(); dis[g][u] = dis[u][g] = min(dis[u][g],w); } for(int k = 1;k <= v;++ k) for(int i = 1;i <= v;++ i) for(int j = 1;j <= v;++ j) if(dis[i][k] + dis[k][j] < dis[i][j]) dis[i][j] = dis[j][i] = dis[i][k] + dis[k][j]; for(int i = 1;i <= n;++ i) for(int j = 0;j <= m;++ j) E[i][j][0] = E[i][j][1] = 999999999; E[1][1][1] = E[1][0][0] = 0; for(int i = 2;i <= n;++ i) { E[i][0][0] = E[i - 1][0][0] + dis[c[i - 1]][c[i]]; for(int j = 1;j <= min(i,m);++ j) { E[i][j][0] = min(E[i - 1][j][1] + k[i - 1] * dis[d[i - 1]][c[i]] + (1 - k[i - 1]) * dis[c[i - 1]][c[i]],E[i - 1][j][0] + dis[c[i - 1]][c[i]]); E[i][j][1] = min(E[i - 1][j - 1][1] + k[i - 1] * k[i] * dis[d[i - 1]][d[i]] + k[i - 1] * (1 - k[i]) * dis[d[i - 1]][c[i]] + (1 - k[i - 1]) * k[i] * dis[c[i - 1]][d[i]] + (1 - k[i - 1]) * (1 - k[i]) * dis[c[i - 1]][c[i]],E[i - 1][j - 1][0] + k[i] * dis[c[i - 1]][d[i]] + (1 - k[i]) * dis[c[i - 1]][c[i]]); } } double minn = 9999999999; for(int i = 0;i <= m;++ i) minn = min(min(E[n][i][1],E[n][i][0]),minn); printf("%.2lf\n", minn); return 0; }
bzoj 2134
因为权值是1,所以此题让求概率 先后俩道题的答案要求一样.
所以就是min(a,b) / a * b.化简一下就是1 / max(a,b)
/*header*/ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <cmath> #define rep(i , x, p) for(int i = x;i <= p;++ i) #define sep(i , x, p) for(int i = x;i >= p;-- i) #define gc getchar() #define pc putchar #define ll long long #define mk make_pair #define fi first #define se second using std::min; using std::max; using std::swap; const int maxN = 10000000 + 7; inline int gi() { int x = 0,f = 1;char c = gc; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘)f = -1;c = gc;} while(c >= ‘0‘ && c <= ‘9‘) {x = x * 10 + c - ‘0‘;c = gc;}return x * f; } void print(int x) { if(x < 0) pc(‘-‘) , x = -x; if(x >= 10) print(x / 10); pc(x % 10 + ‘0‘); } int n , A, B, C, a[maxN]; int main() { scanf("%d%d%d%d%d",&n,&A,&B,&C,a+1); rep(i , 2, n) a[i] = ((long long)a[i-1] * A + B) % 100000001; rep(i , 1, n) a[i] = a[i] % C + 1; double ans = 0; a[n + 1] = a[1]; rep(i , 1, n) ans += 1.0 / max(a[i] , a[i + 1]); printf("%.3lf",ans); return 0; }
luogu 1365
分三种情况考虑贡献即可.
#include <iostream> #include <cstring> #include <cstdio> const int maxN = 300000 + 7; using namespace std; int L,len; char s[maxN]; int main(int argc, char const *argv[]) { scanf("%d",&len); scanf("%s",s + 1); double ans = 0,l = 0; for(int i = 1;i <= len;++ i) { if(s[i] == ‘o‘) {ans += 2.0 * l + 1;l ++;} if(s[i] == ‘x‘) {l = 0;} if(s[i] == ‘?‘) {ans += (2.0 * l + 1) / 2.0;l = (l + 1) / 2;} } printf("%.4lf", ans); return 0; }
luogu3924 康娜的线段树
的确是线段树，比较好做.直接放上std吧，懒得写线段树了
#include<bits/stdc++.h> #define ll long long using namespace std; const int maxn=1e7+5; ll sumv[maxn<<1],a[maxn],s[maxn]; int p[maxn]; namespace io { const int MAXBUF = 1 << 22; char B[MAXBUF], *S = B, *T = B; #define getc() (S == T && (T = (S = B) + fread(B, 1, MAXBUF, stdin), S == T) ? 0 : *S++) template<class Type> inline Type read() { register Type aa = 0; register bool bb = 0; register char ch, *S = io::S, *T = io::T; for(ch = getc(); (ch < ‘0‘ || ch > ‘9‘) && ch != ‘-‘; ch = getc()) ; for(ch == ‘-‘ ? bb = 1 : aa = ch - ‘0‘, ch = getc(); ‘0‘ <= ch && ch <= ‘9‘; ch = getc()) aa = aa * 10 + ch - ‘0‘; io::S = S, io::T = T; return bb ? -aa : aa; } int (*F)() = read<int>; template<> inline double read() { register double aa, bb; register char ch; register char *S = io::S, *T = io::T; while(ch=getc(),(ch<‘0‘||ch>‘9‘)) ;aa=ch-‘0‘; while(ch=getc(),(ch>=‘0‘&&ch<=‘9‘))aa=aa*10+ch-‘0‘; if(ch==‘.‘){bb=1;while(ch=getc(),ch>=‘0‘&&ch<=‘9‘)bb*=0.1,aa+=bb*(ch-‘0‘);} io::S = S, io::T = T; return aa; } char buff[MAXBUF], *iter = buff; template<class T>inline void P(register T x, register char ch = ‘\n‘) { static int stack[110]; register int O = 0; register char *iter = io::iter; if(!x)*iter++ = ‘0‘; else { (x < 0) ? x = -x, *iter++ = ‘-‘ : 1; for(; x; x /= 10) stack[++O] = x % 10; for(; O; *iter++ = ‘0‘ + stack[O--]) ; } *iter++ = ch, io::iter = iter; } inline void putc(register char ch) {*iter++ = ch;} inline void ouput() {fwrite(buff, 1, iter - buff, stdout), iter = buff;} } #define lson o<<1 #define rson o<<1|1 int d=0; inline void build(int l,int r,int o,int t){ if(l==r){sumv[o]=a[l];p[l]=t;d=max(d,t);return;} int mid=l+r>>1; build(l,mid,lson,t+1); build(mid+1,r,rson,t+1); sumv[o]=sumv[lson]+sumv[rson]; } inline ll query(int l,int r,int o,int t,ll tt){ if(l==r){return 1LL*(1LL<<t)*(tt+sumv[o]);} int mid=l+r>>1; return query(l,mid,lson,t-1,tt+sumv[o])+query(mid+1,r,rson,t-1,tt+sumv[o]); } ll gcd(ll a,ll b){ while(b){ ll r=a%b;a=b;b=r; } return a; } int main(){ using io::P; int (*F)()=io::read<int>; int n=F(),m=F(),yyy=F(); for(register int i=1;i<=n;++i)a[i]=F(); build(1,n,1,1); ll ans=query(1,n,1,d-1,0),y=1LL<<(d-1); ll yy=gcd(y,yyy);yyy/=yy;y/=yy; for(register int i=1;i<=n;++i)s[i]=s[i-1]+1LL*(((1LL<<p[i])-1)<<(d-p[i])); while(m--){ int l=F(),r=F(),x=F(); ans+=(s[r]-s[l-1])*x; P(ans/y*yyy); io::ouput(); } }
bzoj 2134
bzoj 2134
constint maxN = 300000 + 7; usingnamespacestd; int L,len; char s[maxN]; intmain(int argc, charconst *argv[]){ scanf("%d",&len); scanf("%s",s + 1); double ans = 0,l = 0; for(int i = 1;i <= len;++ i) { if(s[i] == ‘o‘) {ans += 2.0 * l + 1;l ++;} if(s[i] == ‘x‘) {l = 0;} if(s[i] == ‘?‘) {ans += (2.0 * l + 1) / 2.0;l = (l + 1) / 2;} } printf("%.4lf", ans); return0; }
标签：\n reg rap ios alt data- vector mono space
原文地址：https://www.cnblogs.com/gzygzy/p/10259109.html
