1515377.cpp: 在函数‘int main()’中:
1515377.cpp:13:19: 错误：‘strlen’在此作用域中尚未声明

下边是代码
天大acm第4003题
源码
#include <iostream>
#include <stdio.h>
#define N 100005  
using namespace std;
char x[N];
int num[N];
int main() {
    int t,sum,ans,i;
    scanf("%d",&t);
    while(t--){
        sum=0;
        scanf("%s",x);
        int len=strlen(x);
        for(i=0;i<len;i++){
            num[i]=x[i]-'0';
            sum+=num[i];
        }
        ans=sum%10;
        bool bn=true;
        for(i=0;i<len;i++){
            if(num[i]!=9){
                bn=false;
            }
        }
        int gh=len-1;
        if(len-1==0){
            num[len-1]=10;
            bn=false;
        }
        if(bn==true){   
            num[gh]=9;
            num[0]=10;
            for(i=1;i<len-1;i++){
                num[i]=0;
            }
        }
        if(bn==false){
            if(num[len-1]+10-ans>=10){
                int bit=0;
                bool f=true;
                int u=len-1;
                if(len-1==0){
                    num[len-1]=19;
                    f=false;
                }
                bool hu=true;
                while(f){
                    if(hu){                        
                        if(num[u-1]+1==10){                    
                            bit++;
                        }else{                    
                            sum=sum-num[len-1]+1;               
                            int lk=10-sum%10;
                            num[len-1]=lk;
                            num[u-1]+=1;
                            f=false;
                        }
                        u=u-1;
                        ans=1;
                        hu=false;
                    }else{        
                        if(num[u]+1==10){                        
                            bit++;
                            num[u]=0;
                            u--;
                            ans=1;
                        }else{                    
                            num[u]=num[u]+1;                        
                            sum=sum-(bit-1)*9+1-num[len-1];                        
                            int jh=sum%10;
                            num[len-1]=10-jh;
                            f=false;
                        }                    
                    }

                }

            }else{
                num[len-1] +=10-ans;

            }
        }
        for(int i=0;i<len;i++)  
            printf("%d",num[i]);

        printf("\n");
    }
    system("pause");
    return 0;
}

不行呀

加string.h

天大acm 第4003题，答案有错，提交了的，估计那儿出了问题

The I-number of x is defined to be an integer y, which satisfied the the conditions below:

1. y>x;

2. the sum of each digit of y(under base 10) is the multiple of 10;

3. among all integers that satisfy the two conditions above, y shouble be the minimum.

Given x, you're required to calculate the I-number of x.

Input
An integer T(T≤100) will exist in the first line of input, indicating the number of test cases.

The following T lines describe all the queries, each with a positive integer x. The length of x will not exceed 105.

Output
Output the I-number of x for each query.

Sample Input

1
202



Sample Output

208

感谢各位了，我解决了，不过换了一个方法，代码如下

#include <iostream>
#include <stdio.h>
#include <string.h>
#define N 100005  
using namespace std;
char x[N];
int num[N];
int dum[N];
int solve(int length,int k){
//        printf("k==  %d;length== %d\n",k,length );

    int sum=0;
    num[length]+=k;
    for(int i=length;i>=0;i--){
        if(num[i]>9){
            num[i-1]++;
        }

    }

    for(int j=0;j<=length;j++){
        //            printf("aaalength== %d\n",sum);
        if(j==0){
            if(num[j]==10){
                sum+=1;
            }else{
                sum+=num[j];
            }
        }else{
            sum+=num[j]%10;
            num[j]=num[j]%10;
        }
        

//            printf("sum== %d\n",sum);
//            printf("num[%d]==  %d\n",j,num[j] );
    }
    return sum%10;
}
int main() {
    int t,sum,ans,i;
    scanf("%d",&t);
    while(t--){
        sum=0;
        scanf("%s",x);
        int len=strlen(x);
        for(i=0;i<len;i++){
            num[i]=x[i]-'0';
            dum[i]=x[i]-'0';
            sum+=num[i];
        }
        ans=sum%10;
        int j=num[len-1];
        if(len==1){
            printf("%d",19);

        }else{

            for(i=1;i<20;i++){
                if(solve(len-1,i)==0){
                    for(int j=0;j<len;j++){
                        printf("%d",num[j]);
                    }
               
                    break;
                }else{

                    for(int j=0;j<len;j++){
                        num[j]=dum[j];
                    }
                }

            }
        }
            printf("\n");
    }
//    system("pause");
    return 0;
}

调试时用的，提交时去了的