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题意：给n个数，求最小的段数，使得每一段的最大值之和大于给定的k。每一段的长度相等，最后若干个丢掉。
思路：从小到大枚举段数，如果能o(1)时间求出每一段的和，那么总复杂度是O(n(1+1/2+1/3+...+1/n))=O(nlogn)的。但题目时限卡得比较紧，需加一点小优化，如果连续两个段数它们每一段的个数一样，那么这次只比上次需要多计算一个区间，用上一次的加上这个区间最大值得到当前分段的总和，这样能减少不少运算量。详见代码：
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 #include 10 #include 11 #include 12 #include 13 #include 14 #include 15 #include 16 #include 17 #include 18 #include 19 #include 20 #include 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define lson l, m, rt << 1 26 #define rson m + 1, r, rt << 1 | 1 27 #define define_m int m = (l + r) >> 1 28 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 29 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 30 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 31 #define rep_down1(a, b) for (int a = b; a > 0; a--) 32 #define all(a) (a).begin(), (a).end() 33 #define lowbit(x) ((x) & (-(x))) 34 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 35 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 36 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 37 #define pchr(a) putchar(a) 38 #define pstr(a) printf("%s", a) 39 #define sstr(a) scanf("%s", a); 40 #define sint(a) ReadInt(a) 41 #define sint2(a, b) ReadInt(a);ReadInt(b) 42 #define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c) 43 #define pint(a) WriteInt(a) 44 #define if_else(a, b, c) if (a) { b; } else { c; } 45 #define if_than(a, b) if (a) { b; } 46 #define test_print1(a) cout << "var1 = " << a << endl 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = b" << ", var3 = " << c << endl 49 50 typedef double db; 51 typedef long long LL; 52 typedef pair pii; 53 typedef multiset msi; 54 typedef set si; 55 typedef vector vi; 56 typedef map mii; 57 58 const int dx[8] = { 0, 0, -1, 1}; 59 const int dy[8] = {-1, 1, 0, 0}; 60 const int maxn = 4e5 + 7; 61 const int maxm = 1e3 + 7; 62 const int maxv = 1e7 + 7; 63 const int max_val = 1e6 + 7; 64 const int MD = 22; 65 const int INF = 1e9 + 7; 66 const double pi = acos(-1.0); 67 const double eps = 1e-10; 68 69 template T gcd(T a, T b){ return b==0?a:gcd(b,a%b);} 70 template void ReadInt(T &x){ char c=getchar();while(!isdigit(c))c=getchar();x=0;while(isdigit(c)){x=x*10+c-'0';c=getchar();}} 71 template void WriteInt(T i) { int p=0;static int b[20];if(i == 0) b[p++] = 0;else while(i){b[p++]=i%10;i/=10;}for(int j=p-1;j>=0;j--)pchr('0'+b[j]);} 72 template bool max_update(T &a,const T &b){ if(b>a){a = b; return true;}return false;} 73 template bool min_update(T &a,const T &b){ if(b T condition(bool f, T a, T b){ return f?a:b;} 75 template void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 76 int make_id(int x, int y, int n) { return x * n + y; } 77 78 int f[maxn][20], t[maxn], a[maxn], sum[maxn]; 79 int n; 80 void RMQ_Init() { 81 rep_up0(i, n) f[i][0] = a[i]; 82 rep_up1(j, 18) { 83 for (int i = 0; i + (1 << j) - 1 < n; i++) { 84 f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]); 85 } 86 } 87 } 88 int RMQ(int L, int R) { 89 int p = t[R - L + 1]; 90 return max(f[L][p], f[R - (1 << p) + 1][p]); 91 } 92 LL getSum(int t, int x) { 93 LL sum = 0; 94 rep_up0(i, t) { 95 sum += RMQ(i * x, i * x + x - 1); 96 } 97 return sum; 98 } 99 int main() {100 //freopen("in.txt", "r", stdin);101 //freopen("out.txt", "w", stdout);102 int k;103 rep_up1(i, 18) {104 for (int j = (1 << (i - 1)) + 1; j <= (1 << i); j++) t[j] = i - 1;105 }106 while (cin >> n >> k, n >= 0 || k >= 0) {107 rep_up0(i, n) {108 sint(a[i]);109 if (i) sum[i] = sum[i - 1] + a[i];110 else sum[i] = a[i];111 }112 int ans = -1, last_sum = 0;113 RMQ_Init();114 for (int i = 1; i <= n; i++) {115 int x = n / i;116 LL sum = 0;117 if (i > 1 && n / (i - 1) == x) sum = last_sum + RMQ(x * (i - 1), x * i - 1);118 else sum = getSum(i, x);119 if (sum > k) {120 ans = i;121 break;122 }123 last_sum = sum;124 }125 cout << ans << endl;126 }127 return 0;128 }
转载于:https://www.cnblogs.com/jklongint/p/4427324.html
