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A.疫苗小孩可以三种做法来写:
Sol.1 二分1 1 1针和 0 0 0针的贡献都为 0 0 0,那么固定必打 1 1 1针以上,枚举这一针的位置 p o s pos pos(看作第二针),然后二分找到离 p o s − k pos - k pos−k和 p o s + k pos + k pos+k最近的点,然后讨论计算贡献即可。
注意二分的边界,写丑了就挂了。。。
#include
#define int long long
using namespace std;
const int N = 1e6 + 10;
int a[N], cnt;
string s;
inline void solve(){
int n = 0; cin >> n >> s; s = '@' + s;
for(int i = 1; i > k >> w >> q;
for(int i = 1; i = 1){
if(s[lloc] == '0') lans += w;
else{
int l = 0, r = i - 1;
while(l > 1;
(a[mid] k >> w >> q;
for(int i = 1; i = 1; i--) mr[i] = (s[i] == '0' ? i : mr[i + 1]);
int ans = 0;
for(int i = 1; i i) l = w - abs(abs(i - ml[pos]) - k) * q;
if(mr[pos] && mr[pos] > i) r = w - abs(abs(i - mr[pos]) - k) * q;
ll = max(ll, max(l, r)), pos = max(1ll, i - k), l = -1, r = -1;
if(ml[pos] && ml[pos] k >> w >> q;
for(int i = 1; i = 1; i--) mr[i] = (s[i] == '0' ? i : mr[i + 1]);
for (int i = 1; i
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