Carries
frog has nn integers a1,a2,…,ana1,a2,…,an, and she wants to add them pairwise.
Unfortunately, frog is somehow afraid of carries (进位). She defines hardness h(x,y)h(x,y)for adding xx and yy the number of carries involved in the calculation. For example, h(1,9)=1,h(1,99)=2h(1,9)=1,h(1,99)=2.
Find the total hardness adding nn integers pairwise. In another word, find
∑1≤i<j≤nh(ai,aj)∑1≤i<j≤nh(ai,aj)
.
Input
The input consists of multiple tests. For each test:
The first line contains 11 integer nn (2≤n≤1052≤n≤105). The second line contains nnintegers a1,a2,…,ana1,a2,…,an. (0≤ai≤1090≤ai≤109).
Output
For each test, write 11 integer which denotes the total hardness.
Sample Input
2
5 5
10
0 1 2 3 4 5 6 7 8 9Sample Output
1
20
//题意: n 个数，C(n,2) 这两个数可能进位，问，所有的进位次数是多少?
//题解: 容易想到，枚举可能进位的位置，最多也就9次么，然后二分找可能进位的个数
答案要LL，还wa了一发
1 # include <cstdio> 2 # include <cstring> 3 # include <cstdlib> 4 # include <iostream> 5 # include <vector> 6 # include <queue> 7 # include <stack> 8 # include <map> 9 # include <bitset> 10 # include <sstream> 11 # include <set> 12 # include <cmath> 13 # include <algorithm> 14 #pragma comment(linker,"/STACK:102400000,102400000") 15 using namespace std; 16 #define LL long long 17 #define lowbit(x) ((x)&(-x)) 18 #define PI acos(-1.0) 19 #define INF 0x3f3f3f3f 20 #define eps 1e-8 21 #define MOD 1000000007 22 23 inline int scan() { 24 int x=0,f=1; char ch=getchar(); 25 while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();} 26 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} 27 return x*f; 28 } 29 inline void Out(int a) { 30 if(a<0) {putchar('-'); a=-a;} 31 if(a>=10) Out(a/10); 32 putchar(a%10+'0'); 33 } 34 #define MX 100005 35 /**************************/ 36 int n; 37 int a[MX]; 38 int yu[MX]; 39 40 int bi_search(int l,int r,int w) 41 { 42 int pos = -1; 43 while (l<=r) 44 { 45 int mid = (l+r)/2; 46 if (yu[mid]>=w) 47 { 48 pos = mid; 49 r = mid-1; 50 } 51 else l = mid+1; 52 } 53 return pos; 54 } 55 56 int main() 57 { 58 while (scanf("%d",&n)!=EOF) 59 { 60 int mmm=0; 61 for (int i=1;i<=n;i++) 62 { 63 a[i] =scan(); 64 mmm = max(mmm,a[i]); 65 } 66 LL ans = 0; 67 int y = 1; 68 while(y) 69 { 70 y*=10; 71 for (int i=1;i<=n;i++) 72 yu[i] = a[i]%y; 73 sort(yu+1,yu+1+n); 74 for (int i=1;i<=n;i++) 75 { 76 int pos = bi_search(1,n,y-(a[i]%y)); 77 if (pos!=-1) 78 { 79 int num = n-pos+1; 80 if (a[i]%y<yu[pos]) ans += (LL)num; 81 else ans += (LL)num-1; 82 } 83 } 84 if (y>mmm) break; 85 } 86 printf("%lld ",ans/2); 87 } 88 return 0; 89 }