zoukankan      html  css  js  c++  java
  • HDU6205 card card card

    思维

    把数组扩大一倍,然后找连续的大于0的最长的一段即可,注意结尾特判一下。

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    #define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
    using namespace std;
    typedef long long ll;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int ret = 0, w = 0; char ch = 0;
        while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
        while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
        return w ? -ret : ret;
    }
    inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
    inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
    template <typename T>
    inline T max(T x, T y, T z){ return max(max(x, y), z); }
    template <typename T>
    inline T min(T x, T y, T z){ return min(min(x, y), z); }
    template <typename A, typename B, typename C>
    inline A fpow(A x, B p, C lyd){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
        return ans;
    }
    const int N = 2000005;
    int a[N], b[N], c[N], n;
    
    int main(){
    
        while(~scanf("%d", &n)){
            for(int i = 1; i <= n; i ++) a[i] = read();
            for(int i = 1; i <= n; i ++) b[i] = read();
            for(int i = 1; i <= n; i ++) c[i] = a[i] - b[i];
            for(int i = n + 1; i <= 2 * n; i ++) c[i] = a[i - n] - b[i - n];
            int sum = 0, index = -1, maxi = -INF, ans = -1, len = 0;
            for(int i = 1; i <= 2 * n; i ++){
                if(index == -1 && c[i] < 0) continue;
                else if(index == -1) sum += c[i], index = i, len ++;
                else if(sum + c[i] >= 0){
                    sum += c[i], len ++;
                    if(i == 2 * n){
                        if(len > maxi) maxi = len, ans = index;
                    }
                }
                else if(sum + c[i] < 0){
                    if(len > maxi) maxi = len, ans = index;
                    sum = 0, len = 0, index = -1;
                }
                if(maxi == n) break;
            }
            printf("%d
    ", ans - 1);
        }
        return 0;
    }
    
  • 相关阅读:
    jmeter导出提取的值或参数化的值到excel
    超级有用的正则表达式
    性能测试监控
    asp.net 多线程
    VS代码格式化快捷键
    JS判断IE版本
    Jquery原创排序table
    将aspx转化为html代码
    Java基本语法——变量
    Entity Framework 中的LazyLoading
  • 原文地址:https://www.cnblogs.com/onionQAQ/p/11199236.html
Copyright © 2011-2022 走看看