luogu 1291 概率期望递推

非常好的递推

公式啥的懒得写了,直接放链接哈哈哈https://www.luogu.org/problemnew/solution/P1291

#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i=x;i<=y;i++)
#define dec(i,x,y) for(register int i=x;i>=y;i--)
#define int long long
#define In freopen("7.in","r",stdin)
#define In2 freopen("8.in","r",stdin)
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*f;}
int up,down,s;
inline int gcd(int x,int y){
    while(y){int tmp=x%y;x=y;y=tmp;}return x;}
inline int lcm(int x,int y){
    int gcdd=gcd(x,y);return x*y/gcdd;}
int cal(int x){int s=1;
    for(int i=1;;i++){s=s*10;
    if(s>x) return i;}}
signed main(){
    int n=read();
    if(n==1){puts("1");return 0;}
    if(n==2){puts("3");return 0;}
    
    dec(i,n,1){
        if(!up) {up=1,down=i;continue;}
        int lcmm=lcm(i,down);
        up=lcmm/down*up+lcmm/i;
        down=lcmm;}
    
    up=up*n;
    int gcdd=gcd(up,down);
    up=up/gcdd,down=down/gcdd;
    
    s=up/down;up=up%down;
    
    if(up&&down){
        int k=cal(s),k2=cal(down);
        rep(i,1,k) printf(" ");
        printf("%lld\n",up);
        if(s) printf("%lld",s);
        rep(i,1,k2) printf("-");printf("\n");
        rep(i,1,k) printf(" ");
        printf("%lld\n",down);}
    else printf("%lld",s);  return 0;
}

完结撒花

posted @ 2018-10-15 20:08  ASDIC减除  阅读(110)  评论(0编辑  收藏  举报