题意：给一数字字符串s ( n=s.size()<=1000 ) 和数字m (<1e5) 求s%m
模拟除法， k初值0，按s[0]...累乘相加，把字符串还原成数字，比m大时-m，继续按位还原到s[n-1]
此时剩下的k再%m即为所求
#include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<stack> #include<list> #include<set> using namespace std; typedef long long ll; typedef pair<ll,ll> p; typedef long double ld; #define mem(x) memset(x, 0, sizeof(x)) #define me(x) memset(x, -1, sizeof(x)) #define fo(i,n) for(i=0; i<n; i++) #define sc(x) scanf("%lf", &x) #define pr(x) printf("%lld ", x) #define pri(x) printf("%lld ", x) #define lowbit(x) x&-x const ll MOD = 1e18 +7; const ll N = 6e6 +5; ll a[N], vis[N]; int main() { ll i, j, k, l=0; ll n, m, t, x; string s, s1; while(cin>>s>>m) { k=0;i=0;n=s.size(); while(1) { if(i>=n) break; while(k<m && i<n) { k*=10; k+=s[i++]-'0'; if(k>=m) break; } k-=m; } if(k<0) k+=m; cout<<k%m<<endl; } return 0; }