Ruler method/two points + optimized audition spoj-cran04__ Ruler method

Do you:
1 Knowledge Points: Ruler method/two points + optimization (k = = 0 o'clock to judge the length of the cell and then calculate by formula)

2:
(1): Judge the number of 1 in a 01 sequence with a length of N (n <= 1e6) to determine how many intervals are K
3 Ideas:
(1): ruler (k = = 0 o'clock Special decision)
(2): two points + (k = 0 o'clock Special decision)
4 Reflection:
(1): The ruler of the need to strengthen understanding
(2): Time complexity when critical data (eg:k = = 0) is not taken into account
(3): The number of intervals that need to be stored to satisfy the condition without considering the critical data (eg:k = = 0) by a long long

Vjudge Topic link

The following is accepted code--two points + optimization

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

typedef long Long LL;
int rec[1004014], sum[1004014];

Char str[1004014];
    int main () {LL T, n, K, ans;
    scanf ("%lld", &t);
        while (t--) {scanf ("%lld%lld", &n, &k);
        scanf ("%s", str);
        for (int i = 0; i < n; i++) {rec[i+1] = str[i]-' 0 ';
            } if (!k) {ans = 0;
            LL T, J;
                for (LL i = 1; I <= n; i++) {if (rec[i)) continue;
                for (j = i+1; J <= N; j +) {if (rec[j]) break;
                    } if (j <= N) {t = (j-1)-i+1;
                    Ans + + t* (t+1)/(LL) 2;
                i = j;
                    else {t = n-i+1;
                    Ans + + t* (t+1)/(LL) 2;
                Break
         } printf ("%lld\n", ans);   Continue
        } Sum[0] = 0;
        for (int i = 1; I <= n; i++) {sum[i] = Sum[i-1] + rec[i];
        Ans = 0;
        LL T, L, R, Mid, Mt;
            for (int i = 1; I <= n; i++) {t = Sum[i] + k-rec[i];
            L = i, r = N;
                while (L <= r) {mid = (l+r)/2;
                if (sum[mid] = = t) break;
                if (Sum[mid] < T) {L = mid+1;
                else if (Sum[mid] > t) {r = mid-1;
            } if (L > R) continue;
                else {ans++;
                Mt = mid-1;
                    while (Mt >= i) {if (sum[mt] = = t) ans++;
                    else break;
                mt--;
                Mt = mid+1;
                    while (Mt <= N) {if (sum[mt] = = t) ans++;
                    else break; mt++;
                } printf ("%lld\n", ans);
return 0;
 }

The following is accepted code--ruler + optimization

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

typedef long Long LL;
int rec[1004014];

Char str[1004014];
    int main () {int T, n, K;
    LL ans;
    scanf ("%d", &t);
        while (t--) {scanf ("%d%d", &n, &k);
        scanf ("%s", str);
        for (int i = 1; I <= n; i++) {rec[i] = str[i-1]-' 0 ';
        Ans = 0;
            if (!k) {LL T, J;
                for (LL i = 1; I <= n; i++) {if (rec[i)) continue;
                for (j = i+1; J <= N; j +) {if (rec[j]) break;
                    } if (j <= N) {t = (j-1)-i+1;
                    Ans + + t* (t+1)/(LL) 2;
                i = j;
                    else {t = n-i+1;
                    Ans + + t* (t+1)/(LL) 2;
                Break
     }} else {int id, cnt = 0;       LL A = 0, b = 0;
                    for (int i = 1; I <= n; i++) {if (rec[i] = = 1) {cnt++;
                    if (cnt = = 1) id = i;
                        else if (cnt = = K+1) {ans + = (a+1) * (b+1);
                        A = 0, b = 0, cnt-= 1;
                        id++;
                            while (ID <= n && rec[id] = = 0) {a++;
                        id++;
                    }} else {if (cnt = 0) a++;
                else if (cnt = = k) b++; } if (cnt = = k) ans + = (a+1) * (b+1);/* After the end of the traversal is just CNT cumulative arrival K of the situation needs to consider */} printf ("%lld\n
    ", ans);
return 0;
 }