Starry

发表于 更新于 分类于 本文字数： 508 阅读时长 ≈ 2 分钟

用bitset位操作代替逐个询问

而 已存在，只需 存在，就有 存在

而题意只需要价值存在即可

利用鬼谷子的钱包的分治思想优化解法

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#include <cstdio>
#include <bitset>
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr1(x) printf("Case %d: ", x)
#define pn1(x) printf("Case %d:\n", x)
#define pr2(x) printf("Case #%d: ", x)
#define pn2(x) printf("Case #%d:\n", x)
#define lowbit(x) (x & (-x))

typedef unsigned long long ull;

struct FastIO
{
    static const int S = 5 << 20; //MB
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar()
    {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len)
            return -1;
        return buf[pos++];
    }
    inline int xuint()
    {
        int c = xchar(), x = 0;
        while (~c && c <= 32)
            c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar())
            x = x * 10 + c - '0';
        return x;
    }
    inline LL xull()
    {
        int c = xchar();
        LL x = 0;
        while (~c && c <= 32)
            c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar())
            x = x * 10 + c - '0';
        return x;
    }
    inline int xint()
    {
        int s = 1, c = xchar(), x = 0;
        while (c <= 32)
            c = xchar();
        if (c == '-')
            s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar())
            x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while (c <= 32)
            c = xchar();
        for (; c > 32; c = xchar())
            *s++ = c;
        *s = 0;
    }
    inline void wchar(int x)
    {
        if (wpos == S)
            fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos++] = x;
    }
    inline void wll(LL x)
    {
        if (x < 0)
            wchar('-'), x = -x;

        char s[30];
        int n = 0;
        while (x || !n)
            s[n++] = '0' + x % 10, x /= 10;
        while (n--)
            wchar(s[n]);
    }
    inline void wstring(const char *s)
    {
        while (*s)
            wchar(*s++);
    }
    ~FastIO()
    {
        if (wpos)
            fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;

LL s, sum, a[7], cnt;
bitset<20000> dp;

int main()
{
    //IN;
    // OUT;
    while (++cnt)
    {
        s = 0;
        dp.reset();
        dp.set(0);
        f(i, 1, 6)
            s += (a[i] = io.xint()) * i;
        //io.wll(s);
        if (s == 0)
            break;
        io.wstring("Collection #");
        io.wll(cnt);
        io.wchar(':');
        io.wchar('\n');
        if (s & 1)
            io.wstring("Can't be divided.\n\n");
        else
        {
            s >>= 1;
            f(i, 1, 6)
            {
                for (int j = a[i]; j > 0; j >>= 1)
                {
                    int k = ((j + 1) >> 1) * i;
                    dp |= (dp << k);
                    if (dp[s])
                    {
                        break;
                    }
                }
                if (dp[s])
                    break;
            }
            if (dp[s])
                io.wstring("Can be divided.\n\n");
            else
                io.wstring("Can't be divided.\n\n");
        }
    }
}

发表于 更新于 分类于 本文字数： 172 阅读时长 ≈ 1 分钟

看了题解要分治，一开始打的代码有bug还不清楚什么情况

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#include <cstdio>
#include <algorithm>
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define LL long long
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "w", stdout)
#define scan(x) scanf("%lld", &x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define pr1(x) printf("Case %d: ", x)
#define pn1(x) printf("Case %d:\n", x)
#define pr2(x) printf("Case #%d: ", x)
#define pn2(x) printf("Case #%d:\n", x)
#define lowbit(x) (x & (-x))

typedef unsigned long long ull;
const LL N = 1e4 + 1;
LL m, a[N], cnt;

int main()
{
    scan(m);
    while(m>0){
        a[cnt++]=(m+1)>>1;
        m>>=1;
    }
    sort(a,a+cnt);
    printf("%lld\n",cnt);
    f(i,0,cnt-1)printf("%lld ",a[i]);
    return 0;
}

发表于 更新于 分类于 本文字数： 143 阅读时长 ≈ 1 分钟

今天是大佬带我躺，没想到英语老毛病又犯了height以为是重量

回来时还书不会还，回宿舍路上还撞路灯，脸上开花。。。

不过中午提供的免费盒饭和牛奶倒是可以

一开始想切L，但想法总是有误，欧拉函数没学。。。

而G竟然是泰勒，但好像洛必达也能做，还是太菜了

今天感觉唯一帮助就是个双目三目运算符，。。。。。。

该肝了

感谢队友不杀之恩！

发表于 更新于 分类于 本文字数： 186 阅读时长 ≈ 1 分钟 
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#include<bits/stdc++.h>
#define scan(x) scanf("%lld",&x)
#define f(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long LL;
const LL N=186;
LL dp[N][N],data[N][N],r,c,maxn;
LL dyp(LL i,LL j) {
    if(dp[i][j]!=-1) return dp[i][j];
    LL a[4]= {1,1,1,1};
    if(i-1>0&&data[i-1][j]<data[i][j]) {
        a[0]+=dyp(i-1,j);
    }
    if(i+1<=r&&data[i+1][j]<data[i][j]) {
        a[1]+=dyp(i+1,j);
    }
    if(j-1>0&&data[i][j-1]<data[i][j]) {
        a[2]+=dyp(i,j-1);
    }
    if(j+1<=c&&data[i][j+1]<data[i][j]) {
        a[3]+=dyp(i,j+1);
    }
    sort(a,a+4);
    dp[i][j]=a[3];
    return a[3];
}

int main() {
    scan(r);
    scan(c);
    f(i,1,r) {
        f(j,1,c) {
            scan(data[i][j]);
            dp[i][j]=-1;
        }
    }
    f(i,1,r) {
        f(j,1,c) {
            maxn=maxn<dyp(i,j)?dyp(i,j):maxn;
        }
    }

    printf("%lld",maxn);
    return 0;
}

发表于 更新于 分类于 本文字数： 409 阅读时长 ≈ 1 分钟

有一个样例是全为0三角形，而一开始dp数组未初始化为特殊值仅特判是否为0，从而导致TLE

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#include <iostream>
#include <cstdio>
#define scan(x) scanf("%lld",&x)
#define f(i,a,b) for(int i=a;i<=b;i++)
#define pn(x,y) printf("###%lld:::%lld\n",x,y)
using namespace std;
typedef long long LL;
const LL N=1e3+1;
LL a[N][N],r,f[N][N];

struct FastIO {
static const int S = 5 << 20; //MB
int wpos; char wbuf[S];
FastIO() : wpos(0) {}
inline int xchar() {
static char buf[S];
static int len = 0, pos = 0;
if(pos == len)
pos = 0, len = fread(buf, 1, S, stdin);
if(pos == len) return -1;
return buf[pos ++];
}
inline int xuint() {
int c = xchar(), x = 0;
while(~c && c <= 32) c = xchar();
for(; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x;
}
inline LL xull() {
int c = xchar();
LL x = 0;
while(~c && c <= 32) c = xchar();
for(; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x;
}
inline int xint() {
int s = 1, c = xchar(), x = 0;
while(c <= 32) c = xchar();
if(c == '-') s = -1, c = xchar();
for(; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x * s;
}
inline void xstring(char* s) {
int c = xchar();
while(c <= 32) c = xchar();
for(; c > 32; c = xchar()) * s++ = c;
*s = 0;
}
inline void wchar(int x) {
if(wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
wbuf[wpos ++] = x;
}
inline void wll(LL x) {
if(x < 0) wchar('-'), x = -x;

char s[30];
int n = 0;
while(x || !n) s[n ++] = '0' + x % 10, x /= 10;
while(n--) wchar(s[n]);
}
inline void wstring(const char* s) {
while(*s) wchar(*s++);
}
~FastIO() {
if(wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
}
} io;

LL dyp(LL i,LL j) {
    if(i==r)return a[i-1][j];
    if(f[i][j]!=-1)return f[i][j];
    LL q=a[i-1][j]+(dyp(i+1,j)<dyp(i+1,j+1)?dyp(i+1,j+1):dyp(i+1,j));
    f[i][j]=q;
    return q;
}
int main() {
    r=io.xint();
    f(i,1,r) {
        f(j,1,i) {
            a[i-1][j-1]=io.xint();
            f[i][j-1]=-1;
        }
    }
    io.wll(dyp(1,0));
    return 0;
}
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