Codeforces Round #353 (Div. 2) A. Infinite Sequence

题目链接：

http://codeforces.com/contest/675/problem/A

题意：

给你一个等差数列，问你b是不是其中的一项。

即是否存在大于等于0的整数k使得a+k*c==b。只要求(b-a)%c==0&&(b-a)/c>=0就可以了。

但是，我就是写成了二分。。而且还写搓了。。

先看看哇的代码：

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

const int maxn = 333;
const int INF = 1e9;

int main() {
    int a, b, c;
    while (scanf("%d%d%d", &a, &b, &c) == 3) {
        if (c == 0) {
            if (b == a) puts("YES");
            else puts("NO");
            continue;
        }
        int low, hig,flag;
        if (c < 0) {
            flag = 0;
            low=0, hig = (-INF - a) / c + 10;
            if (b > a) {
                puts("NO"); continue;
            }
            while (low < hig-1) {
                int mid = low + (hig - low) / 2;
                if (a + mid*c>b) {
                    low = mid + 1;
                }
                else if (a + mid*c < b) {
                    hig = mid;
                }
                else {
                    flag = 1; break;
                }
            }
            if (flag) puts("YES");
            else puts("NO");
        }
        else if (c>0) {
            flag = 0;
            low = 0, hig = (INF - a) / c + 10;
            if (b < a) {
                puts("NO"); continue;
            }
            while (low < hig - 1) {
                int mid = low + (hig - low) / 2;
                if (a + mid*c<b) {
                    low = mid + 1;
                }
                else if (a + mid*c > b) {
                    hig = mid;
                }
                else {
                    flag = 1; break;
                }
            }
            if (flag) puts("YES");
            else puts("NO");
        }
    }
    return 0;
}

以后写二分以此为鉴。。

贴修改之后的：

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

const int maxn = 333;
typedef long long LL;
const LL INF = 1e9;

int main() {
    int a, b, c;
    while (scanf("%d%d%d", &a, &b, &c) == 3) {
        if (c == 0) {
            if (b == a) puts("YES");
            else puts("NO");
            continue;
        }
        LL low, hig;
        if (c < 0) {
            low = 0, hig = (-INF - a) / c + 10;
            if (b > a) {
                puts("NO"); continue;
            }
            while (low < hig - 1) {
                LL mid = low + (hig - low) / 2;
                if (a + mid*c>=b) {
                    low = mid;
                }
                else if (a + mid*c < b) {
                    hig = mid;
                }
            }
            if (a + low*c == b) puts("YES");
            else puts("NO");
        }
        else if (c>0) {
            low = 0, hig = (INF - a) / c + 10;
            if (b < a) {
                puts("NO"); continue;
            }
            while (low < hig - 1) {
                LL mid = low + (hig - low) / 2;
                if (a + mid*c<=b) {
                    low = mid;
                }
                else if (a + mid*c > b) {
                    hig = mid;
                }
            }
            if (a + low*c == b) puts("YES");
            else puts("NO");
        }
    }
    return 0;
}