本文共 1913 字，大约阅读时间需要 6 分钟。
https://www.luogu.org/problemnew/show/P4942
上面为原题地址；
引理：
一个数字除以9的余数等于它的各位数字之和除以9的余数
那么我们将其各个数字加起来%9即可；
当然用等差数列公式更快；
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include //#include //#pragma GCC optimize("O3")using namespace std;#define maxn 400005#define inf 0x3f3f3f3f#define INF 9999999999#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9 + 7;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-3typedef pair pii;#define pi acos(-1.0)const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair pii;inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}ll sqr(ll x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/ll qpow(ll a, ll b, ll c) { ll ans = 1; a = a % c; while (b) { if (b % 2)ans = ans * a%c; b /= 2; a = a * a%c; } return ans;}int T;ll l, r, cnt;int main(){ //ios::sync_with_stdio(0); rdint(T); while (T--) { rdllt(l); rdllt(r); cnt = (r - l + 1); if (cnt % 2 == 0) { cout << ((cnt / 2) % 9 * (l + r) % 9) % 9 << endl; } else if ((r + l) % 2 == 0) { cout << ((cnt) % 9 * ((l + r) / 2) % 9) % 9 << endl; } } return 0;}
转载于:https://www.cnblogs.com/zxyqzy/p/9966931.html