NYOJ 873 环之最大和

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题目意思很简单就是让你从n个数字中选取m个连续的数字使和最大就是这样没错因为是环状的所以可以从末尾连到最开始

PS：如果你觉得写得一定没有错但还是wa 你可能忽略了 long long 没错恭喜你！！！（还要注意最后 MOD n）

方法一：可以用前缀和遍历一遍得到最大值和边界问题

#include <iostream>#include <algorithm>#include <cstring>using namespace std;const int maxn = 3000;int main(){int n, m, arr[maxn];long long pres[maxn];while (cin >> n >> m) {memset(pres , 0, sizeof(pres));for (int i = 1; i <= n; i ++) {cin >> arr[i];pres[i] = pres[i - 1] + arr[i];}for (int i = n + 1; i <= n + m; i ++) {arr[i] = arr[i - n];pres[i] = pres[i - 1] + arr[i];}//for (int i = 1; i <= n + m - 1; i ++) {//cout << arr[i] << " ";//}cout << endl;//for(int i = 1; i <= n + m - 1; i ++) {//cout << pres[i] << " ";//}int head , tail;long long ans = -999999;for (int i = m; i <= n + m; i ++) {if(pres[i] - pres[i - m ] > ans) {ans = pres[i] - pres[i - m ];head = i - m + 1;tail = i;}}if(head > n)head %= n;if(tail > n)tail %= n;cout << ans << " " << head << " " << tail << endl;}}

方法二是这样的可以用尺取法维护两个伪指针和中间变量来得到最大值这样的好处在于如果数组特别特别大前缀和无法保存的时候而且也比前缀和的方法更快

#include <iostream>using namespace std;const int maxn = 3000;int main(){int n, m, arr[3000];while (cin >> n >> m) {for (int i = 1 ;i <= n; i ++) {cin >> arr[i];}for (int i = n + 1; i <= n + m - 1; i ++) {arr[i] = arr[i - n];}int head = 1, tail = m;long long ans = -999999;long long temp = 0;for (int i = 1; i <= m; i ++) temp += arr[i];ans = temp;int l = head;int r = tail;for (int i = 1; i < n; i ++) {temp += arr[++r];temp -= arr[l++];if( temp > ans ) {ans = temp;head = l;tail = r;}}if(head > n)head %= n;if(tail > n)tail %= n;cout << ans << " " << head << " " << tail << endl;}}

没错我把两种方法对应反了你以为我是不小心的？？笑话我怎么可能没发现我就是故意的就是这么傲娇

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