hz18790581821的博客

#include #include #include #include #define Mod 99991using namespace std;long long int ans,x,y;void fun(long long int n){ if(n==0) { ans=1; return; } if(n==1) { ans=x%Mod; return;

#include #include #include #include using namespace std;void fun(char s[],int c[]) //字符串转逆序数组 { for(int i=0;i<strlen(s);i++) c[i]=s[strlen(s)-1-i]-'0';}int main(){ char str1[105],str2[1005

大明A+BTime Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11953    Accepted Submission(s): 4350Problem Description话说，经过了漫长的一个多月，小明已经成

#include #include #include #include using namespace std;struct sz{ //结构体定义 int d[1000]; int len; sz(){ //初始化结构体 memset(d,0,sizeof(d)); len =0;

#include #include #include #include using namespace std;int a[50000];int main(){ int n; while(~scanf("%d",&n)) { memset(a,0,sizeof(a)); a[0]=1; for(int i=1;i<=n;i++) { int c=0;

#include #include #include #include using namespace std;void fun(char s[],int c[]){ for(int i=0;i<strlen(s);i++) { c[i]=s[strlen(s)-1-i]-'0'; }}int main(){ char s1[15],s2[15];

DescriptionIn many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given