题目传送门：https://codeforces.com/problemset/problem/1181/B
题目大意： 给定一串长为 $ l(l\leqslant10^5) $ 的数字，可将数字分成无前导零的两个数\(A,B\)，问\(A+B\)的最小值
将数字分成长度相等的两部分显然是最优的，我们只需要判断是否有前置零，将断点前移或后移判断即可
/*program from Wolfycz*/ #include #include #include #include #include #include #include #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair #define int_inf 0x7f7f7f7f using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } templateinline T frd(T x){ int f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } templateinline T read(T x){ int f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int digit=8; const int base=1e8; const int maxn=2e4; char s[maxn*digit],temp[maxn*digit]; struct Bignum{ int V[maxn],len; Bignum(){len=1,memset(V,0,sizeof(V));} void read(char *s){ int l=strlen(s),tim=1; reverse(s,s+l); len=(l-1)/digit+1; for (int i=0;i>1;pos>0;pos--){ if (temp[pos]=='0') continue; Bignum A; A.read(temp+pos); temp[pos]='\0'; Bignum B; B.read(temp); if (A+B>1)+1;pos