Sha 10458-Cricket Ranking)

Sha 10458-Cricket Ranking)
Ultraviolet A 10458-Cricket Ranking

Question Link

Given k intervals, you must use these numeric ranges to combine them into n.

Idea: the concept of the review is as follows: if n components are known to be s and there is no limit to the number of values, use the partition method to find the situation C (s + n-1, n-1), but this part contains more than, so we can use the binary to list which are beyond, so that s minus f (I) + 1 in this way, the position is exceeded. After this part is subtracted, more Subtraction is added back. This satisfies the formula of the refresh principle. If the number is an odd number, the value is subtracted, returns an even number.

Code:

#include 
 
  #include 
  
   #include 
   
    using namespace std;typedef long long ll;const int MAXN = 1005;struct bign {    int len;    ll num[MAXN];    bign () {len = 0;memset(num, 0, sizeof(num));    }    bign (ll number) {*this = number;}    bign (const char* number) {*this = number;}    void DelZero ();    void Put ();    void operator = (ll number);    void operator = (char* number);    bool operator <  (const bign& b) const;    bool operator >  (const bign& b) const { return b < *this; }    bool operator <= (const bign& b) const { return !(b < *this); }    bool operator >= (const bign& b) const { return !(*this < b); }    bool operator != (const bign& b) const { return b < *this || *this < b;}    bool operator == (const bign& b) const { return !(b != *this); }    void operator ++ ();    void operator -- ();    bign operator + (const int& b);    bign operator + (const bign& b);    bign operator - (const int& b);    bign operator - (const bign& b);    bign operator * (const ll& b);    bign operator * (const bign& b);    bign operator / (const ll& b);    //bign operator / (const bign& b);    int operator % (const int& b);};/*Code*/int k;long long n, f[10];int bitcount(int x) {    return x == 0 ? 0 : bitcount(x>>1) + (x&1);}bign C(long long n, long long m) {    bign ans = 1;    for (long long i = 0; i < m; i++)ans = ans * (n - i) / (i + 1);    return ans;}int main() {    while (~scanf("%d%lld", &k, &n)) {long long l, r;for (int i = 0; i < k; i++) {    scanf("%lld%lld", &l, &r);    f[i] = r - l;    n -= l;}bign ans1 = 0LL, ans2 = 0LL;for (int i = 0; i < (1<
    
     = 0; i--) printf("%lld", num[i]);}void bign::operator = (char* number) {    len = strlen (number);    for (int i = 0; i < len; i++)num[i] = number[len-i-1] - '0';    DelZero ();}void bign::operator = (ll number) {    len = 0;    while (number) {num[len++] = number%10;number /= 10;    }    DelZero ();}bool bign::operator < (const bign& b) const {    if (len != b.len)return len < b.len;    for (int i = len-1; i >= 0; i--)if (num[i] != b.num[i])    return num[i] < b.num[i];    return false;}void bign::operator ++ () {    int s = 1;    for (int i = 0; i < len; i++) {s = s + num[i];num[i] = s % 10;s /= 10;if (!s) break;    }    while (s) {num[len++] = s%10;s /= 10;    }}void bign::operator -- () {    if (num[0] == 0 && len == 1) return;    int s = -1;    for (int i = 0; i < len; i++) {s = s + num[i];num[i] = (s + 10) % 10;if (s >= 0) break;    }    DelZero ();}bign bign::operator + (const int& b) {    bign a = b;    return *this + a;}bign bign::operator + (const bign& b) {    int bignSum = 0;    bign ans;    for (int i = 0; i < len || i < b.len; i++) {if (i < len) bignSum += num[i];if (i < b.len) bignSum += b.num[i];ans.num[ans.len++] = bignSum % 10;bignSum /= 10;    }    while (bignSum) {ans.num[ans.len++] = bignSum % 10;bignSum /= 10;    }    return ans;}bign bign::operator - (const int& b) {    bign a = b;    return *this - a;}bign bign::operator - (const bign& b) {    ll bignSub = 0;    bign ans;    for (int i = 0; i < len || i < b.len; i++) {bignSub += num[i];if (i < b.len)    bignSub -= b.num[i];ans.num[ans.len++] = (bignSub + 10) % 10;if (bignSub < 0) bignSub = -1;else bignSub = 0;    }    ans.DelZero();    return ans;}bign bign::operator * (const ll& b) {    ll bignSum = 0;    bign ans;    ans.len = len;    for (int i = 0; i < len; i++) {bignSum += num[i] * b;ans.num[i] = bignSum % 10;bignSum /= 10;    }    while (bignSum) {ans.num[ans.len++] = bignSum % 10;bignSum /= 10;    }    return ans;}bign bign::operator * (const bign& b) {    bign ans;    ans.len = 0;     for (int i = 0; i < len; i++){  int bignSum = 0;  for (int j = 0; j < b.len; j++){      bignSum += num[i] * b.num[j] + ans.num[i+j];      ans.num[i+j] = bignSum % 10;      bignSum /= 10;}  ans.len = i + b.len;  while (bignSum){      ans.num[ans.len++] = bignSum % 10;      bignSum /= 10;}      }      return ans;}bign bign::operator / (const ll& b) {    bign ans;    ll s = 0;    for (int i = len-1; i >= 0; i--) {s = s * 10 + num[i];ans.num[i] = s/b;s %= b;    }    ans.len = len;    ans.DelZero();    return ans;}int bign::operator % (const int& b) {    bign ans;    int s = 0;    for (int i = len-1; i >= 0; i--) {s = s * 10 + num[i];ans.num[i] = s/b;s %= b;    }    return s;}