1 条题解
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0
C :
#include <stdio.h> int main() { int m,n,a[20000],carry,i,j,digit = 1,temp; a[0] = 1; scanf("%d %d",&m,&n); for(i = 1; i <= n; ++i) { carry = 0; for(j = 1; j <= digit; ++j) { temp = a[j-1] * m + carry; a[j-1] = temp % 10; carry = temp / 10; } while(carry) { a[++digit-1] = carry % 10; carry /= 10; } } for(i = digit; i >=1; --i) { printf("%d",a[i-1]); } return 0; }
C++ :
#include<iostream> using namespace std; unsigned R[20000] = {0}; int main() { int i,j; int X,Y; unsigned temp; unsigned temp1; unsigned temp2 = 1; cin >> X >> Y; R[0] = 1; for(i = 1; i <= Y; i++) { temp = 0; for(j = 1; j <= temp2; j++) { temp1 = R[j-1] * X + temp; R[j-1] = temp1 % 10; temp = temp1 / 10; } while(temp) { R[++temp2-1] = temp % 10; temp /= 10; } } for(i = temp2;i >= 1; i--) { cout << R[i-1]; } return 0; }
- 1
信息
- ID
- 1348
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者