题目描述
CC is a smart girl and she is curious about everything. One day she starts to analyze her lifestyle and finds out that she either feels happy or unhappy in any day, then she collects her mood data and makes numerical representation. She writes down 1 if she is happy, otherwise she writes down 0.
For example, if CC is happy, happy, unhappy, happy, happy, unhappy in the next 6 days, a numerical sequence which stands for it is 1,1,0,1,1,0
Meanwhile, smart CC finds her mood sequence is always periodic. She also finds the repeated segment like 1,1,0 in the aforementioned example and calls it “feeling repetend”. As a curious girl, CC wants to make another deep analysis. After she gets the length of “feeling repetend” n, she defines a rule: she tries to find a day whose index is d and considers the next d+n-1 days. For each day d‘ in [d, d+n-1], CC guarantee that the number of happy days is more than unhappy days in [1, d‘] days.
Now, CC wonder that how many days which satisfy the rule in n “feeling repetend” days.
输入描述:
Input contains multiple test cases.The first line contains an integer T (1<=T<=20), which is the number of test cases.Then the first line of each test case contains an integer n (1<=n<=100000), which is the length of “feeling repetend”.The second line of each test case contains n integers, each integer is either 0 or 1, standing for unhappy or happy.
输出描述:
output a integer represents the answer.
示例1
输入
2 5 1 0 1 1 0 5 1 1 1 1 1
输出
1 5
说明
For the sample,the days from the third day as a starting point are as follows:Day1: 1 day happy : 0 days unhappyDay2: 2 days happy : 0 days unhappyDay3: 2 days happy : 1 day unhappyDay4: 3 days happy : 1 day unhappyDay5: 3 days happy : 2 days unhappy
题解
$O(n)$扫描。
题意比较难懂，而且题意中有描述错的东西，看了样例才渐渐明白。
题意：有一个长度为$2*n$的数组，输入了$a[1]$到$a[n]$，剩下的是前面的拷贝。问有多少个$x$满足$x$属于$[1, n]$，且在区间$[x, x+n-1]$中，$1$的个数大于$0$的个数。
维护两个值，$1$的个数和$0$的个数，每次区间往后面移动一下，两个值更新一下，看有多少符合要求的就可以了。
赛中比较蠢蛋，弄了个前缀和在搞，不过也是可以的。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 10;
int T, n;
int a[maxn], sum[maxn];
int m[4 * maxn];
void build(int l, int r, int rt) {
if(l == r) {
m[rt] = sum[l];
return;
}
int mid = (l + r) / 2;
build(l, mid, 2 * rt);
build(mid + 1, r, 2 * rt + 1);
m[rt] = min(m[2 * rt], m[2 * rt + 1]);
}
int get(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
return m[rt];
}
int mid = (l + r) / 2;
int left = 300000;
int right = 300000;
if(L <= mid) left = get(L, R, l, mid, 2 * rt);
if(R > mid) right = get(L, R, mid + 1, r, 2 * rt + 1);
return min(left, right);
}
int main() {
scanf("%d", &T);
while(T --) {
scanf("%d", &n);
for(int i = 1; i <= n; i ++) {
scanf("%d", &a[i]);
if(a[i] == 0) a[i] --;
a[n + i] = a[i];
}
for(int i = 1; i <= 2 * n; i ++) {
sum[i] = sum[i - 1] + a[i];
}
build(1, 2 * n, 1);
int ans = 0;
for(int i = 1; i <= n; i ++) {
if(get(i, i + n - 1, 1, 2 * n, 1) <= sum[i - 1]) {
continue;
}
ans ++;
}
printf("%d\n", ans);
}
return 0;
}