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Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expressionnext(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the%I64d specificator.
2 7
33
7 7
7
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
让后对区间分段处理即可
code:
#include#include #include #include #include using namespace std;typedef long long LL;#define maxn 1023LL dx[] = {4,7};int cnt ;LL a[maxn];struct node{ LL x;};void bfs(){ a[0] = 4;a[1] = 7; cnt = 2; queue Q; node p,q; p.x = 4;Q.push(p); p.x = 7;Q.push(p); while(!Q.empty()){ q = Q.front();Q.pop(); for(int i = 0;i < 2;i++) { p.x = q.x * 10 + dx[i]; a[cnt ++] = p.x; Q.push(p); if(p.x > 1e9) return; } }}int main(){ bfs(); LL r,l,n,m; while(cin >> l >> r){ LL sum = 0; while(l <= r){ int k = lower_bound(a,a+maxn,l) - a; if(l == a[k] ) {l++;sum += a[k];} else { n = r - l; m = a[k] - l; if(n >= m) sum += a[k] * (m+1); else sum += a[k] * (n + 1); l = a[k] + 1; } } cout << sum << endl; }}
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