2015多校第九场 HDU 5400 Arithmetic Sequence 数学

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5400

题意：定义(d1,d2)算术序列为：对于给定的序列b1,b2,...,bn，存在i使得bj+1=bj+d1( j∈[1,i) );bj+1=bj+d2（j∈[i,n)）.现在给出d1,d2和一个序列a1,a2,...,an，找出有多少个区间[l,r]满足（d1,d2）算术序列。

解法：我们可以找出对于给定的序列，满足算术序列的所有最大子串的长度，然后对于每一个子串，找出他的所有子串即可。对于一个长度为n的序列，其所有非空子串的个数为n(n+1)/2（可以由挡板原理算出）。加了读入挂竟然跑到了HDU上面的第一名。

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 100010;int n, x, y, d1, d2, sub[maxn];struct FastIO{    static const int S = 1310720;    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) exit(0);        return buf[pos ++];    }    inline int xuint()    {        int c = xchar(), x = 0;        while (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 wint(LL x)    {        if (x < 0) wchar('-'), x = -x;        char s[24];        int n = 0;        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;        while (n--) wchar(s[n]);        wchar('\n');    }    inline void wstring(const char *s)    {        while (*s) wchar(*s++);    }    ~FastIO()    {        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;    }} io;int main(){    while(1)    {        n = io.xint();        d1 = io.xint();        d2 = io.xint();        y = io.xint();        for(int i=1; i<n; i++){            x = y;            y=io.xint();            sub[i] = y-x;        }        sub[n] = 0x3f3f3f3f;        bool flag = 1;        LL ans = n;        int l = 0, r = 0;        for(int i=1; i<=n; i++)        {            if(sub[i] == d1 && flag){                l++;            }            else if(sub[i] == d2){                r++;                flag = false;            }            else{                ans += 1LL*(l+r)*(l+r+1)/2;                l = r = 0;                flag = true;                if(sub[i] == d1){                    l++;                }            }        }        io.wint(ans);    }    return 0;}

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